life cycle costing for engineersviii contents 3.3 life cycle costing application areas 28 3.4 life...

Life Cycle Costingfor Engineers

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Life Cycle Costingfor Engineers

B.S. DHILLON

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CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

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Library of Congress Cataloging-in-Publication Data

Dhillon, B. S. (Balbir S.), 1947-Life cycle costing for engineers / author, editor, B.S. Dhillon.

p. cm.“A CRC title.”Includes bibliographical references and index.ISBN 978-1-4398-1688-2 (hard back : alk. paper)1. Life cycle costing. 2. Engineering economy. 3. Product life cycle. I. Title.

TA177.7.D3525 2010658.15’52--dc22 2009030894

Visit the Taylor & Francis Web site athttp://www.taylorandfrancis.com

and the CRC Press Web site athttp://www.crcpress.com

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This book is affectionately dedicated to my dear friend,

Dr. G. S. Guram, in thanks for his guidance, honesty, support,

and friendship over the years.

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vii

Contents

Preface ................................................................................................................... xiiiThe Author .......................................................................................................... xvii

1 Introduction .....................................................................................................11.1 Background ............................................................................................11.2 Terms and Definitions ..........................................................................21.3 Useful Information on Life Cycle Costing ........................................3

1.3.1 Journals .....................................................................................31.3.2 Conference Proceedings .........................................................41.3.3 Technical Reports and Manuals ............................................41.3.4 Books..........................................................................................51.3.5 Data Information Sources .......................................................61.3.6 Organizations ...........................................................................6

1.4 Scope of the Book ..................................................................................7Problems ............................................................................................................8References .........................................................................................................9

2 Life Cycle Costing Economics .................................................................... 112.1 Introduction ......................................................................................... 112.2 Simple Interest ..................................................................................... 112.3 Compound Interest ............................................................................. 122.4 Effective Annual Interest Rate .......................................................... 142.5 Time-Dependent Formulas for Application in Life Cycle

Cost Analysis ....................................................................................... 152.5.1 Single Payment Future Worth Formula .............................. 152.5.2 Single Payment Present Value Formula .............................. 152.5.3 Uniform Periodic Payment Future Amount Formula ...... 162.5.4 Uniform Periodic Payment Present Value Formula .......... 182.5.5 Formulas to Calculate Value of Annuity Payments

When Annuity’s Present and Future Values Are Given .... 192.6 Depreciation Methods ........................................................................ 20

2.6.1 Sum-of-Years-Digits (SYD) Method ..................................... 212.6.2 Straight-Line Method ............................................................222.6.3 Declining-Balance Method ...................................................22

Problems .......................................................................................................... 24References .......................................................................................................25

3 Life Cycle Costing Fundamentals ............................................................. 273.1 Introduction ......................................................................................... 273.2 Need and Information Required for Life Cycle Costing ............... 27

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viii Contents

3.3 Life Cycle Costing Application Areas ..............................................283.4 Life Cycle Costing Activities and Associated Steps ...................... 293.5 Approach for Incorporating Life Cycle Costing into the

Planning Process for Proposals and Contracts ...............................303.6 Areas for Evaluating a Life Cycle Costing Program ...................... 313.7 Life Cycle Costing Data Sources ....................................................... 323.8 Life Cycle Costing Advantages and Disadvantages

and Related Important Points ...........................................................333.9 Life Cycle Costing Concept Application in Selecting

Equipment from Competing Manufacturers ..................................34Problems ..........................................................................................................40References ....................................................................................................... 41

4 Life Cycle Cost Models and Cost Estimation Methods ........................434.1 Introduction .........................................................................................434.2 Types of Life Cycle Cost Models and Their Inputs ........................434.3 General Life Cycle Cost Models ........................................................44

4.3.1 General Life Cycle Cost Model I ..........................................444.3.2 General Life Cycle Cost Model II .........................................454.3.3 General Life Cycle Cost Model III .......................................464.3.4 General Life Cycle Cost Model IV ....................................... 474.3.5 General Life Cycle Cost Model V ........................................ 474.3.6 General Life Cycle Cost Model VI ....................................... 49

4.4 Specific Life Cycle Cost Models ........................................................504.4.1 Specific Life Cycle Cost Model I ..........................................504.4.2 Specific Life Cycle Cost Model II ......................................... 514.4.3 Specific Life Cycle Cost Model III ........................................ 524.4.4 Specific Life Cycle Cost Model IV .......................................534.4.5 Specific Life Cycle Cost Model V .........................................55

4.5 Cost Estimation Methods ...................................................................554.5.1 Cost Estimation Method I .....................................................554.5.2 Cost Estimation Method II ...................................................564.5.3 Cost Estimation Method III .................................................. 574.5.4 Cost Estimation Method IV .................................................. 574.5.5 Cost Estimation Method V ...................................................58

Problems .......................................................................................................... 59References ....................................................................................................... 59

5 Reliability, Quality, Safety, and Manufacturing Costing ....................635.1 Introduction .........................................................................................635.2 Reliability Cost Classifications ..........................................................635.3 Models for Estimating Costs of Reliability-Related Tasks ............64

5.3.1 Model I .....................................................................................645.3.2 Model II ...................................................................................65

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Contents ix

5.3.3 Model III ..................................................................................655.3.4 Model IV ..................................................................................655.3.5 Model V ...................................................................................66

5.4 Quality Cost Classifications and Their Distribution in the Industrial Sector .......................................................................665.4.1 Prevention Cost ...................................................................... 675.4.2 Appraisal Cost ........................................................................ 675.4.3 Internal Failure Cost .............................................................. 675.4.4 External Failure Cost ............................................................. 67

5.5 Quality Cost Indexes and Quality Cost Reduction Approach ..............................................................................................68

5.6 Safety Cost and Its Related Facts and Figures ................................ 695.7 Safety Cost Estimation Models ......................................................... 70

5.7.1 Model I ..................................................................................... 705.7.2 Model II ................................................................................... 705.7.3 Model III .................................................................................. 715.7.4 Model IV .................................................................................. 71

5.8 Manufacturing Costs ..........................................................................725.9 Manufacturing Cost Estimation Models ......................................... 73

5.9.1 Model I ..................................................................................... 735.9.2 Model II ................................................................................... 735.9.3 Model III .................................................................................. 745.9.4 Model IV .................................................................................. 74

Problems ..........................................................................................................75References .......................................................................................................75

6 Maintenance, Maintainability, Usability, and Warranty Costing .................................................................................776.1 Introduction .........................................................................................776.2 Reasons for Maintenance Costing, Factors Influencing

Maintenance Cost, and Types of Maintenance Costs .................... 786.3 Equipment Maintenance Cost ........................................................... 79

6.3.1 Maintenance Equipment Cost ..............................................806.4 Preventive and Corrective Maintenance Labor Cost

Estimation ............................................................................................806.5 Repair Manpower, Maintenance Material, and Spare

and Repair Parts Costs ....................................................................... 816.6 Maintenance Cost Estimation Models .............................................83

6.6.1 Model I .....................................................................................836.6.2 Model II ...................................................................................846.6.3 Model III ..................................................................................846.6.4 Model IV ..................................................................................84

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x Contents

6.7 Maintenance Cost Data Collection ...................................................856.8 Maintainability Investment Cost Elements .....................................856.9 Manufacturer Warranty and Reliability Improvement

Warranty Costs ....................................................................................866.10 Usability Costing and Related Facts and Figures .......................... 876.11 Principal Costs of Ignoring Product Usability

and Product Usability Cost Estimation ........................................... 87Problems .......................................................................................................... 89References ....................................................................................................... 89

7 Computer System Life Cycle Costing ....................................................... 917.1 Introduction ......................................................................................... 917.2 Computer System Life Cycle Cost Models ...................................... 917.3 Computer System Maintenance Cost ............................................... 937.4 Software Costing and Related Difficulties ...................................... 957.5 Software Life Cycle Cost Influencing Factors and Model ............. 967.6 Software Cost Estimation Methods and Models ............................ 97

7.6.1 Software Cost Estimation Methods .................................... 977.6.1.1 Tabular Models ....................................................... 987.6.1.2 Composite Models .................................................. 987.6.1.3 Analytic Models .....................................................997.6.1.4 Linear Models .........................................................997.6.1.5 Multiplicative Models ............................................99

7.6.2 Software Cost Estimation Models ..................................... 100Problems ........................................................................................................ 103References ..................................................................................................... 103

8 Transportation System Life Cycle Costing ........................................... 1058.1 Introduction ....................................................................................... 1058.2 Aircraft Life Cycle Cost .................................................................... 1058.3 Aircraft Turbine Engine Life Cycle Cost ....................................... 1088.4 Aircraft Cost Drivers ........................................................................ 108

8.4.1 Helicopter Maintenance Cost Drivers .............................. 1098.4.2 Aircraft Airframe Maintenance Cost Drivers ................. 1098.4.3 Combat Aircraft Hydraulic and Fuel Systems

Cost Drivers .......................................................................... 1108.5 Cargo Ship Life Cycle Cost .............................................................. 1108.6 Operating and Support Costs for Ships ......................................... 111

8.6.1 Formula I ............................................................................... 1118.6.2 Formula II .............................................................................. 1118.6.3 Formula III ............................................................................ 1118.6.4 Formula IV ............................................................................ 112

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Contents xi

8.7 Urban Rail Life Cycle Cost ............................................................... 1128.8 Car Life Cycle Cost ........................................................................... 1138.9 City Bus Life Cycle Cost Estimation Model .................................. 114Problems ........................................................................................................ 115References ..................................................................................................... 115

9 Civil Engineering Structures and Energy Systems Life Cycle Costing ...................................................................................... 1179.1 Introduction ....................................................................................... 1179.2 Building Life Cycle Cost .................................................................. 1179.3 Steel Structure Life Cycle Cost ........................................................ 1189.4 Bridge and Waste Treatment Facilities Life Cycle Costs ............. 1199.5 Building Energy Cost Estimation ................................................... 120

9.5.1 Formula I ............................................................................... 1209.5.2 Formula II .............................................................................. 1219.5.3 Formula III ............................................................................ 1219.5.4 Formula IV ............................................................................ 1229.5.5 Formula V ............................................................................. 122

9.6 Appliance Life Cycle Costing .......................................................... 1229.7 Energy System Life Cycle Cost Estimation Model ....................... 1239.8 Motor, Pump, and Circuit-Breaker Life Cycle Costs .................... 124

9.8.1 Motor Life Cycle Cost Estimation Model ......................... 1249.8.2 Pump Life Cycle Cost Estimation Model ......................... 1259.8.3 Circuit-Breaker Life Cycle Cost Estimation Model ......... 126

Problems ........................................................................................................ 126References ..................................................................................................... 127

10 Miscellaneous Cost Estimation Models ................................................ 12910.1 Introduction ....................................................................................... 12910.2 Plant Cost Estimation Model ........................................................... 12910.3 Reliability Acquisition Cost Estimation Model ............................ 13010.4 Development Cost Estimation Model ............................................ 13110.5 Program Error Cost Estimation Model .......................................... 13210.6 Cooling Tower Cost Estimation Model .......................................... 13310.7 Storage Tank Cost Estimation Model ............................................. 13410.8 Pressure Vessel Cost Estimation Model......................................... 13410.9 New Aircraft System Spares Cost Estimation Model .................. 13610.10 Satellite Procurement Cost Estimation Model .............................. 13710.11 Single-Satellite System Launch Cost Estimation Model .............. 13710.12 Tank Gun System Life Cycle Cost Estimation Model .................. 13810.13 Weather Radar Life Cycle Cost Estimation Model ....................... 139Problems ........................................................................................................ 141References ..................................................................................................... 142

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xii Contents

11 Introduction to Engineering Reliability and Maintainability ......... 14511.1 Introduction ....................................................................................... 14511.2 Reliability and Maintainability Definitions .................................. 14611.3 Bathtub Hazard Rate Curve ............................................................ 14611.4 General Reliability, Mean Time to Failure,

and Hazard Rate Formulas .............................................................. 14711.4.1 General Formula for Reliability ......................................... 14711.4.2 General Formula for Mean Time to Failure ..................... 14811.4.3 General Formula for Hazard Rate ..................................... 149

11.5 Common Reliability Networks ....................................................... 15011.5.1 Series Network ..................................................................... 15011.5.2 Parallel Network .................................................................. 15211.5.3 K-out-of-m Network ............................................................. 15411.5.4 Standby System .................................................................... 155

11.6 Reliability and Maintainability Relationship ............................... 15611.6.1 Reliability .............................................................................. 15611.6.2 Maintainability ..................................................................... 157

11.7 Maintainability Measures ................................................................ 15711.7.1 Mean Time to Repair (MTTR) ............................................ 15711.7.2 Maintainability Function .................................................... 15811.7.3 Mean Preventive Maintenance Time ................................ 15911.7.4 Maximum Corrective Maintenance Time ........................ 159

11.7.4.1 Exponential ........................................................... 15911.7.4.2 Normal ................................................................... 16011.7.4.3 Lognormal ............................................................. 160

11.8 System Availability and Unavailability ......................................... 16011.9 Reliability and Maintainability Tools ............................................ 162

11.9.1 Failure Modes and Effect Analysis (FMEA) .................... 16211.9.2 Fault Tree Analysis .............................................................. 16211.9.3 Cause and Effect Diagram .................................................. 163

Problems ........................................................................................................ 164References ..................................................................................................... 165

Bibliography: Literature on Life Cycle Costing .......................................... 167

Index ..................................................................................................................... 197

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xiii

Preface

Today, in the global economy, the procurement decisions for many engi-neering products, particularly expensive ones, are not made on initial pro-curement costs alone, but rather on their life cycle costs. Past experiences indicate that often product ownership cost exceeds the procurement cost. In fact, according to some studies, the product ownership cost (i.e., logistics and operating cost) can vary from 10 to 100 times the original acquisition cost.

Over the past 20 years, a large number of journal and conference proceed-ings articles on life cycle costing have appeared; however, to my knowledge, only two or three books specifically covering certain areas of civil engineer-ing have been published. More specifically, no general book on life cycle cost-ing was published during this period. In 1989, I published a general book on the topic by reviewing and listing all the journal and conference proceedings articles up to 1989.

The absence of an up-to-date general book on the topic has caused a great deal of difficulty for information seekers because they have had to consult many different and diverse sources. Thus, the main objective of this book is to cover all the latest and most useful aspects of life cycle costing in a single volume and thus eliminate the need to consult many different and diverse sources to obtain desired information. The sources of most of the material presented are listed in the reference section at the end of each chapter. These will be useful to readers who desire to delve more deeply into a specific area or topic.

The book contains a chapter on life cycle costing economics and another on introductory engineering reliability and maintainability concepts consid-ered useful to understanding other chapters of the book. The topics covered in the book are treated in such a manner that the reader does not need previ-ous knowledge to understand the contents. At appropriate places, the book contains examples, along with their solutions; at the end of each chapter, numerous problems test reader comprehension. An extensive list of publica-tions on life cycle costing covering the period from 1988 to 2008 is provided in the bibliography at the end of this book to give readers a view of the inten-sity of developments on the topic.

The book is composed of 11 chapters. Chapter 1 presents a historical back-ground of life cycle costing, frequently used terms and definition in life cycle costing, useful information on life cycle costing, and the scope of the book. Chapter 2 is devoted to economics concepts considered useful to perform life cycle cost analysis; it also covers topics such as simple interest, compound interest, effective annual interest rate, time-dependent economics formulas, and depreciation methods.

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xiv Preface

Chapter 3 presents various aspects of life cycle costing fundamentals, including the need and information required for life cycle costing, life cycle costing application areas, approach for incorporating life cycle costing into the planning process for proposals and contracts, areas for evaluating the life cycle costing program, life cycle costing advantages and disadvantages, and life cycle costing data sources. A number of life cycle cost models and cost estimation methods are covered in Chapter 4. The life cycle cost models in the chapter are divided into two areas: general and specific.

Chapter 5 is devoted to reliability, quality, safety, and manufacturing costing. Some of the topics covered in the chapter are reliability cost clas-sifications, models for estimating the cost of reliability-related tasks, qual-ity cost classifications, quality cost indexes, safety cost and its related facts and figures, safety cost estimation models, and manufacturing cost estima-tion models. Chapter 6 presents various important aspects of maintenance, maintainability, usability, and warranty costing. It covers topics such as rea-sons for maintenance costing, factors influencing maintenance cost, types of maintenance costs, preventive and corrective maintenance labor cost estima-tion, maintenance cost data collection, maintainability investment cost ele-ments, manufacturer warranty and reliability improvement warranty costs, usability costing and related facts and figures, and principal costs of ignor-ing product usability and product usability cost estimation.

Chapters 7 and 8 are devoted to computer system life cycle costing and transportation system life cycle costing, respectively. Some of the topics cov-ered in Chapter 7 are computer system life cycle cost models, computer sys-tem maintenance cost, software life cycle cost influencing factors and model, and software cost estimation methods and models. Chapter 8 includes top-ics such as aircraft life cycle cost, aircraft turbine engine life cycle cost, air-craft cost drivers, cargo ship life cycle cost, ship operating and support costs, urban rail life cycle cost, and city bus life cycle cost estimation models.

Chapter 9 presents various important aspects of civil engineering struc-tures and energy systems life cycle costing. Some of the topics covered in the chapter are building life cycle cost, steel structure life cycle cost, bridge and waste treatment facilities life cycle costs, building energy cost estimation, appliance life cycle costing, and an energy system life cost estimation model. Chapter 10 is devoted to miscellaneous cost estimation models and it pres-ents a total of 12 such models. Some of these models include the plant cost estimation model, program error cost estimation model, satellite procure-ment cost estimation model, and tank gun system life cycle cost estimation model.

Finally, Chapter 11 presents various introductory aspects of engineering reliability and maintainability. The topics covered in the chapter include bathtub hazard rate curve; common reliability networks; general reliability, mean time to failure, and hazard rate formulas; maintainability measures; and reliability and maintainability tools.

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Preface xv

This book will be useful to many individuals, including engineering professionals at large, engineering undergraduate and graduate students, engineering administrators, cost analysts, engineering researchers and instructors, and procurement professionals.

I am deeply indebted to many individuals, including colleagues, students, and friends, for their input and encouragement throughout the project. I thank my children, Jasmine and Mark, for their patience, as well as intermit-tent disturbances that resulted in many desirable breaks! Last, but not least, I thank my boss, other half, and wife, Rosy, for typing various portions of this book and for her timely help in proofreading and tolerance.

B. S. DhillonOttawa, Ontario

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xvii

The Author

B. S. Dhillon is a professor of engineering management in the Department of Mechanical Engineering at the University of Ottawa. He has served as a chairman/director of the Mechanical Engineering Department/Engineering Management Program for over 10 years at the same institution. He has pub-lished 343 articles (201 journal articles and 142 conference proceedings) on reliability, safety, engineering management, etc. He is or has been on the editorial boards of nine international scientific journals. In addition, Dr. Dhillon has written 35 books on various aspects of reliability, design, safety, quality, and engineering management published by Wiley (1981), Van Nostrand (1982), Butterworth (1983), Marcel Dekker (1984), Pergamon Press (1986), etc. His books are being used in over 100 countries and many of them have been translated into languages such as German, Russian, and Chinese. He served as general chairman of two international conferences on reliabil-ity and quality control held in Los Angeles and Paris in 1987.

Professor Dhillon has served as a consultant to various organizations and bodies and has many years of experience in the industrial sector. At the University of Ottawa, he has taught reliability, quality, engineering man-agement, design, and related areas for over 29 years. He has also lectured in over 50 countries, including keynote addresses at various international scientific conferences held in North America, Europe, Asia, and Africa. In March 2004, Dr. Dhillon was a distinguished speaker at the Conference/Workshop on Surgical Errors (sponsored by the White House Health and Safety Committee and the Pentagon) held on Capitol Hill.

Professor Dhillon attended the University of Wales, where he received a BS degree in electrical and electronic engineering and an MS degree in mechanical engineering. He received a PhD degree in industrial engineering from the University of Windsor.

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1

1Introduction

1.1 Background

Today, in the global economy and due to various other market pressures, the acquisition decisions of many engineering systems, particularly the expen-sive ones, are not made based on initial procurement costs but rather on their life cycle costs. Past experiences indicate that often engineering system own-ership costs exceed acquisition costs. In fact, according to various studies [1], the engineering system ownership cost (i.e., logistic and operating cost) can vary from 10 to 100 times the original acquisition cost.

The life cycle cost of a system may be defined simply as the sum of all costs incurred during its life span (i.e., the total of acquisition and ownership costs). The term life cycle costing was used for the first time in 1965 in a report entitled “Life Cycle Costing in Equipment Procurement” [2]. This report was prepared by the Logistics Management Institute, Washington, D.C., for the assistant secretary of defense for installations and logistics, U.S. Department of Defense, Washington, D.C.

As a result of this document, the Department of Defense published a series of three guidelines for life cycle costing procurement, entitled “Life Cycle Costing Procurement Guide (Interim),” “Life Cycle Costing in Equipment Procurement—Casebook,” and “Life Cycle Costing Guide for System Acquisitions (Interim)” [3–5]. In 1971, the Department of Defense issued Directive 5000.1, entitled “Acquisition of Major Defense Systems,” concern-ing the requirement for life cycle costing procurement for major systems acquisitions [6].

In 1974, the concept of life cycle costing was formally adopted by the state of Florida and, in 1975, a project entitled “Life Cycle Budgeting and Costing as an Aid in Decision Making” was initiated by the Untied States Department of Health, Education, and Welfare [7]. In 1978, the U.S. Congress passed the National Energy Conservation Policy Act, which made it manda-tory for every new federal building to be life cycle cost effective [8].

Since 1974, states such as New Mexico, Alaska, Maryland, North Carolina, and Texas have passed legislation that make life cycle cost analysis manda-tory in the planning, design, and construction of all state buildings [8]. In 1981, a journal article presented a comprehensive list of publications on life cycle costing [9]. In 1989, Dhillon presented a list of over 500 publications on various aspects of life cycle costing [8].

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2 Life Cycle Costing for Engineers

Since 1989, many people have contributed to the subject of life cycle cost-ing. An extensive list of publications on life cycle costing covering 1988–2007 is presented in the bibliography at the end of this book.

1.2 Terms and Definitions

Many terms and definitions are used in the area of life cycle costing. Some of the frequently used terms and definitions that are directly or indirectly related to life cycle costing include [8,10–15]:

Cost• is the amount of money paid or payable for the acquirement of materials, property, or services.Procurement cost• is the total of investment or acquisition costs (non-recurring and recurring).Ownership cost• is the total of all costs other than the procurement cost during the life span of an item.Life cycle cost• is the sum of all costs incurred during the life span of an item or system (i.e., the total of procurement and ownership costs).Recurring cost• is the cost that recurs periodically during the life span of a project or item.Nonrecurring cost• is the cost that is not repeated.Reliability• is the probability that an item or system will perform its function satisfactorily for the desired period when used according to specified conditions.Maintainability• is the probability that a failed item or system will be restored to its satisfactory working state within a stated total downtime when maintenance action is started per specified conditions.Downtime• is the total time during which the item or system is not in a condition to perform its specified mission or function.Manufacturing cost• is the sum of fixed and variable costs chargeable to the manufacture of a specified item or system.Maintenance• is all scheduled and unscheduled actions necessary to keep an item or system in a serviceable state or restore it to ser-viceability. It includes inspection, servicing, modification, repair, etc.Repair cost• is the cost of restoring an item, system, or facility to its original performance or condition.

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Introduction 3

Maintenance cost• is the materials and labor expense required for maintaining items in satisfactory use condition.Mean time to repair• is the average or mean time required to repair an item or system.Failure• is the termination of the ability of an item or system to per-form its specified function or mission.Failure rate• is the number of failures of an item or system per unit measure of life (e.g., hours).Compound amount• is the future value of money loaned or invested at compound interest.Redundancy• is the existence of more than one means to perform a specified function.Annuity• is a series of equal payments at equal time intervals.Cost model• is an approach, based on programmatic and technical parameters, for calculating concerned costs.Cost estimating relationship• is an equation that relates cost as the dependent variable to one or more independent variables.Useful life• is the length of time an item or system functions within an acceptable level of failure rate.Mission time• is the time during which the item or system is carrying out its stated mission.

1.3 Useful Information on Life Cycle Costing

There are many sources for obtaining, directly or indirectly, life cycle costing–related information. Some of the most useful sources are listed under the following various different categories.

1.3.1 Journals

IEEE Transactions on Reliability•Information and Management•Journal of Quality in Maintenance Engineering•International Power Generation•Microelectronics and Reliability•Better Roads•Journal of Infrastructure Systems•International Journal of Production Research•

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Railway Gazette International•Concrete Engineering International•IEEE Aerospace and Electronic Systems Magazine•Journal of Transportation Engineering•International Journal of Quality and Reliability Management•Defense Management Journal•Transportation Research Record•International Journal of Production Economics•Chemical Engineering•Quality Engineering•IEEE Transactions on Power Delivery•Reliability Engineering and System Safety•Rail International•European Transactions on Electric Power•

1.3.2 Conference Proceedings

Proceedings of the Annual Reliability and Maintainability Symposium•Proceedings of the Annual ISSAT International Conference on Reliability •and Quality in Design

Proceedings of the Annual Reliability Engineering Conference for the •Electric Power Industry

Proceedings of the Annual American Society for Quality Control •(ASQC) Conference

Proceedings of the IEEE Annual Conference on Industrial Electronics•Proceedings of the Annual Offshore Technology Conference•Proceedings of the Annual Canadian Society for Civil Engineering Conference•Proceedings of the Annual Petroleum and Chemical Industry Conference•Proceedings of the IEEE Annual Pulp and Paper Industry Technical •Conference

Proceedings of the Annual Conference of the Urban and Regional Information •Systems Association

1.3.3 Technical Reports and Manuals

MIL-HDBK-259 (Navy), “Life Cycle Cost in Navy Acquisitions,” •Department of Defense, Washington, D.C., April 1983MIL-HDBK-276-1 (MC), “Life Cycle Cost Model for Defense Material •Systems Data Collection Workbook,” Department of Defense, Washington, D.C., February 1984

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Introduction 5

NIST Handbook 135, “Life Cycle Costing Manual: For the Federal •Energy Management Program,” U.S. Department of Energy, Washington, D.C., 1995NISTIR 6806, “Project-Oriented Life Cycle Costing Workshop for •Energy Conservation in Buildings,” U.S. Department of Energy, Washington, D.C., September 2001NISTIR-85-3273-21 (Rev. 4/06), “Energy Price Indices and Discount •Factors for Life Cycle Cost Analysis,” U.S. Department of Energy, Washington, D.C., April 2006“Life Cycle Cost Analysis: A Guide for Architects,” American •Institute of Architects, Washington, D.C., 1977D. E. Peterson, “Life Cycle Cost Analysis of Pavements,” Trans-•portation Research Board, National Research Council, Washington, D.C., 1985H. Hawk, “Bridge Life Cycle Cost Analysis,” Transportation Research •Board, National Research Council, Washington, D.C., 2003D. M. Frangopol and H. Furuta, editors, “Life Cycle Cost Analysis •and Design of Civil Infrastructure Systems,” Structural Engineering Institute of the American Society of Civil Engineers, Reston, VA, 2001

1.3.4 Books

B. S. Blanchard, • Design and Manage to Life Cycle Cost, M/A Press, Portland, OR, 1978A. Boussabaine and R. Kirkham, • Whole Life Cycle Costing, Blackwell Publishing, Oxford, UK, 2004J. W. Bull, editor, • Life Cycle Costing for Construction, Blackie Academic and Professional, Inc., London, 1993W. J. Fabrycky and B. S. Blanchard, • Life Cycle Cost and Economic Analysis, Prentice Hall, Inc., Englewood Cliffs, NJ, 1991B. S. Dhillon, • Life Cycle Costing: Techniques, Models, and Applications, Gordon and Breach Science Publishers, New York, 1989D. Hunkeler, K. Lichtenvort, and G. Rebitzer, editors, • Environmental Life Cycle Costing, CRC Press, Boca Raton, FL, 2008M. R. Seldon, • Life Cycle Costing: A Better Method of Government Procurement, Westview Press, Boulder, CO, 1979A. J. Dell’isola and S. J. Kirk, • Life Cycle Costing for Design Professionals, McGraw–Hill Book Company, New York, 1981M. E. Earles, • Factors, Formulas, and Structures for Life Cycle Costing, Eddins–Earles, Concord, MA, 1981R. J. Brown and R. R. Yanuck, • Life Cycle Costing: A Practical Guide for Energy Managers, Fairmont Press, Inc., Atlanta, GA, 1980

K10869_Book.indb 5 8/26/09 1:44:26 PM


1.3.5 Data Information Sources

Government Industry Data Exchange Program (GIDEP)•GIDEP Operations CenterNaval Weapons StationU.S. Department of NavySeal BeachCorona, CA 91720

National Technical Information Center (NTIS)•5285 Port Royal RoadSpringfield, VA 22151

Defense Technical Information Center•DTIC-FDAC8725 John J. Kingman Road, Suite 0944Fort Belvoir, VA 22060-6218

Reliability Analysis Center•Rome Air Development CenterGriffiss Air Force BaseRome, NY 13441-5700

American National Standards Institute (ANSI)•11 W. 42nd StreetNew York, NY 10036

Technical Services Department•American Society for Quality611 W. Wisconsin AvenueP.O. Box 3005Milwaukee, WI 53201-3005

1.3.6 Organizations

American Society of Civil Engineers (ASCE)•1801 Alexander Bell DriveReston, VA 20190-4400.

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Introduction 7

Society of Manufacturing Engineers•One SME DriveDearborn, MI 48121.

American Society of Mechanical Engineers (ASME)•Three Park AvenueNew York, NY 10016-5990.

American Society of Heating, Refrigeration and Air Conditioning •Engineers (ASHRAE)1791 Tullie Circle, NEAtlanta, GA 30329.

American Public Power Association•1875 Connecticut Avenue, NW, Suite 1200Washington, D.C. 20009-5715.

SOLE—The International Society of Logistics•8100 Professional Place, Suite 111Hyattsville, MD 20785-2229.

Reliability Society, IEEE•P.O. Box 1331Piscataway, NJ

1.4 Scope of the Book

Nowadays, life cycle costing is receiving increasing attention in various sec-tors of the economy, including government procurements and industry. Over the past two decades, a large number of journal and conference proceedings articles have appeared; however, to the best of the author’s knowledge, only two or three books specifically covering certain areas of civil engineering have been published. More specifically, no general book on life cycle costing has been produced during this period.

Professionals and others involved in life cycle costing need up-to-date information on the subject and generally face a great deal of difficulty

K10869_Book.indb 7 8/26/09 1:44:27 PM


because they have to consult many different and diverse sources. This book is an attempt to satisfy this specific need. The book is written after review-ing the currently available literature on life cycle costing. Therefore, all the effort was directed to covering important past and present issues in the field.

Previous knowledge is not generally required to understand the mate-rial covered in this book because two chapters on fundamental eco-nomics and reliability and maintainability basics are provided to give sufficient background to potential readers. The book will find use in many diverse disciplines and will be useful to engineering professionals at large, engineering undergraduate and graduate students, procurement professionals, engineering instructors and researchers, and engineering administrators.

Problems

1. Write an essay on the historical developments in life cycle costing.

2. List at least five sources for obtaining life cycle costing–related information.

3. List at least five books considered important for obtaining life cycle costing-related information.

4. Define the following three terms:life cycle cost•ownership cost•nonrecurring cost•

5. List three of the most important organizations for obtaining life cycle costing-related information.

6. List five important journals from which to obtain life cycle costing-related information.

7. Define the following terms:repair cost•maintenance cost•procurement cost•

8. What is the difference between the terms maintainability and maintenance?

9. Compare the meanings of the following terms:recurring cost•nonrecurring cost•

10. What is the difference between the terms equipment reliability and equipment maintainability?

K10869_Book.indb 8 8/26/09 1:44:27 PM

Introduction 9

References

1. Ryan, W. J. 1968. Procurement views of life cycle costing. Proceedings of the Annual Symposium on Reliability 164–168.

2. Logistics Management Institute (LMI). 1965. Life cycle costing in equipment procurement. Report no. LMI task 4C-5, Washington, D.C.

3. U.S. Department of Defense. 1970. Life cycle costing procurement guide (interim). Department of Defense guide no. LCC-1, Washington, D.C.

4. U.S. Department of Defense. 1970. Life cycle costing in equipment procure-ment—Casebook. Department of Defense guide no. LCC–2, Washington, D.C.

5. U.S. Department of Defense. 1973. Life cycle costing guide for system acquisi-tions (interim). Department of Defense guide no. LCC–3, Washington, D.C.

6. U.S. Department of Defense. 1971. Acquisition of major defense systems. Department of Defense directive no. 5000.1, Washington, D.C.

7. Earles, M. E. 1978. Factors, formulas, and structures for life cycle costing. Concord, MA: Eddins–Earles.

8. Dhillon, B. S. 1989. Life cycle costing: Techniques, models, and applications. New York: Gordon and Breach Science Publishers.

9. Dhillon, B. S. 1981. Life cycle cost: A survey. Microelectronics and Reliability 21:495–511.

10. Humphreys, K. K., and Wellman, P. 1987. Basic cost engineering. New York: Marcel Dekker, Inc.

11. Stewart, R. D., and Wyskida, R. M. 1987. Cost estimator’s reference manual. New York: John Wiley & Sons.

12. Humphreys, K. K. 1984. Project and cost engineers’ handbook. New York: Marcel Dekker, Inc.

13. Society of Automotive Engineers, Inc. 1987. Aircraft engine life cycle cost. Document no. SP-721, Warrendale, PA.

14. Seldon, M. R. 1979. Life cycle costing: A better method of government procurement. Boulder, CO: Westview Press.

15. Brown, R. J., and Yanuck, R. R. 1980. Life cycle costing. Atlanta, GA: The Fairmount Press, Inc.

K10869_Book.indb 9 8/26/09 1:44:27 PM

K10869_Book.indb 10 8/26/09 1:44:27 PM

11

2Life Cycle Costing Economics

2.1 Introduction

The discipline of economics plays a key role in life cycle costing because, to calculate the life cycle cost of items, various types of economics-related infor-mation are required. Life cycle costing requires that all potential costs be calculated by taking into consideration the time value of money. In modern society, interest and inflation rates are utilized to take into consideration the time value of money.

In fact, the concept of interest is not new; its history may be traced back to 2000 BC in Babylon, where interest on borrowed commodities (e.g., grain) was paid in the form of grain or through other possible means [1]. Thus, in a similar manner in modern times, the future value of present dollars will be greater because of earned interest or smaller because of inflation. Similarly, the present value of an amount of money to be received in the future would generally be less.

In life cycle costing, future costs, such as operation and maintenance costs associated with an item, have to be discounted to their present values before adding them to the item’s acquisition or procurement cost. Over the years, many formulas have been developed in the area of economics for convert-ing money from one point of time to another. Such formulas are considered indispensable in life cycle costing.

This chapter presents various aspects of economics considered useful in performing life cycle costing studies.

2.2 Simple Interest

This is the simplest form of interest and it means that the interest is paid only on the original amount of money borrowed, rather than on the accrued interest. Thus, the total interest paid on the borrowed amount of money is expressed by

I P n i= ( )( )( ) (2.1)

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whereI is total interest.P is principal amount (i.e., borrowed).n is total number of interest periods (e.g., years).i is interest rate per specified period.

The total amount of money, A, at the end of, say, n years is expressed by

A P I= + (2.2)

By substituting Equation (2.1) into Equation (2.2), we get

A P P n i

P ni

= +

= +

( )( )( )

( )1 (2.3)

Example 2.1A company borrowed $300,000 for a period of 3 years at an annual simple inter-est rate of 5% to procure engineering equipment. Calculate the total amount of money the company has to pay to the lender at the end of 3 years.

By substituting the given data values into Equation (2.3), we get

A = +

=

( , ) ( ( . ) ( ))

$ ,

300 000 1 0 05 3

345 000

Thus, the total amount of money the company has to pay to the lender at the end of 3 years is $345,000.

2.3 Compound Interest

In this case, the interest earned on principal amount, P, during each inter-est period is added (at the end of each period) to the principal amount and thereafter begins earning interest itself for the remaining term of the loan or investment. Thus, at the end of the first interest period (e.g., a year) the total amount is expressed by

A P P i

P i

1

1

= +

= +

( )( )

( ) (2.4)

where A1 is the total amount at the end of the first interest period.

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Life Cycle Costing Economics 13

At the end of the second interest period (e.g., a year), the total amount is expressed by

A A i2 1 1= +( ) (2.5)

By substituting Equation (2.4) into Equation (2.5), we obtain

A P i i

P i

2

2

1 1

1

= + +

= +

( ) ( )

( ) (2.6)

where A2 is the total amount at the end of the second interest period.Similarly, at the end of the third interest period (e.g., a year), the total

amount is expressed by

A A i3 2 1= +( ) (2.7)


A P i i

P i

32

3

1 1

1

= + +

= +

( ) ( )

( ) (2.8)

where A3 is the total amount at the end of the third interest period.Thus, at the end of the nth interest period (e.g., a year), the total amount is

generalized to the following form:

A A i

P i

n n

n

= +

= +

−1 1

1

( )

( ) (2.9)

wheren is number of interest periods (e.g., years).An is total or compound amount at the end of the nth interest period (e.g.,

a year).An–1 is principal amount at the beginning of the nth interest period (e.g.,

a year).

The total compound interest earned after the nth interest period (e.g., a year) is given by

I A Pc n= − (2.10)

K10869_Book.indb 13 8/26/09 1:45:25 PM


Example 2.2Assume that a person deposited $80,000 in a bank for 7 years at annual interest rate of 7%, compounded annually. Calculate the total amount of money at the end of the specified period and the compound interest earned at the end of the same period.

By substituting the given data values into Equations (2.9) and (2.10), we get

A7780 000 1 0 07

128 462 52

= +

=

( , ) ( . )

$ , .

and

Ic = + −

=

( , ) ( . ) ( , )

$ , .

80 000 1 0 07 80 000

48 462 52

7

Thus, the total amount of money and the compound interest earned at the end of 7 years are $128,462.52 and $48,462.52, respectively.

2.4 Effective Annual Interest Rate

This interest rate may be described simply as the true annual interest rate because it considers the effect of all compounding during the year. The effective annual interest rate can be calculated by using the following equa-tion [2]:

( )1 1+ = +

iime

m

(2.11)

whereie is effective annual interest rate.i is annual nominal interest rate.m is total number of interest periods in a year.

Note that Equation (2.11) is developed by reasoning that the effective inter-est rate compounded once a year generates the same interest as a nominal interest rate compounded m times in a year. By rearranging Equation (2.11), we get

i

ime

m

= +

−1 1 (2.12)

K10869_Book.indb 14 8/26/09 1:45:46 PM


Example 2.3A person deposited $100,000 in a bank at a nominal interest rate of 8% com-pounded monthly, for 12 months. Estimate the value of the effective annual inter-est rate.

By substituting the specified data values into Equation (2.12), we get

ie = +

−

= −

=

=

10 0812

1

1 08299 1

0 08299

8 299

12.

.

.

. %

Thus, the value of the effective annual interest rate is 8.299%.

2.5 Time-Dependent Formulas for Application in Life Cycle Cost Analysis

In the published literature, many time-dependent formulas have been devel-oped that can be used to perform life cycle cost analysis. Some of these for-mulas are presented next.

2.5.1 Single Payment Future Worth Formula

This formula for compound amount was developed earlier in the chapter (in the section on compound interest). Thus, from Equation (2.9), the future worth (compound amount) is

W A P if nn= = +( )1 (2.13)

whereWf is future worth or amount (i.e., principal amount plus interest earned).n is number of interest periods (e.g., years).P is principal amount. i is compound interest rate per specified period.

2.5.2 Single Payment Present Value Formula

From Equation (2.13), the present value of a future amount of money is given by

V P

Wipf

n= =

+( )1 (2.14)

where Vp is the present value.

K10869_Book.indb 15 8/26/09 1:46:03 PM


Example 2.4Assume that the total operation and maintenance cost of a piece of engineering equipment after its 5-year usage will be $150,000. Calculate the present value of $150,000 if the annual compound interest rate is 6%.


Vp =+

=

( , )( . )

$ , .

150 0001 0 06

112 088 7

5

Thus, the present value of the engineering equipment total operation and mainte-nance cost is $112,088.70.

2.5.3 Uniform Periodic Payment Future Amount Formula

This formula is concerned with determining the future amount at the end of n interest periods (years) of equal payments made at the end of each interest period. All payments are invested at an annual compound interest rate i. The formula is developed next.

At the end of the first year, after the first payment, the future amount is

FA PA1 = (2.15)

whereFA1 is future amount at the end of the first year.PA is payment made at the end of a year.

At the end of the second year, after the second payment and the interest earned on FA1, using Equation (2.4) the future amount is

FA PA FA i2 1 1= + +( ) (2.16)

whereFA2 is future amount at the end of the second year.i is annual compound interest rate.

Substituting Equation (2.15) into Equation (2.16) yields

FA PA PA i2 1= + +( ) (2.17)

At the end of the third year, after the third payment and the interest earned on FA2, the future amount is

FA PA FA i3 2 1= + +( ) (2.18)

where FA3 is the future amount at the end of the third year.

K10869_Book.indb 16 8/26/09 1:46:30 PM



FA PA PA i PA i321 1= + + + +( ) ( ) (2.19)

At the end of the fourth year, after the fourth payment and the interest earned on FA3, the future amount is

FA PA FA i4 3 1= + +( ) (2.20)

where FA4 is the future amount at the end of the fourth year.Using Equation (2.19) in Equation (2.20) yields

FA PA PA i PA i PA i42 31 1 1= + + + + + +( ) ( ) ( ) (2.21)

At the end of the nth year, after the nth payment and the interest earned on FAn–1, the future amount is

FA PA PA i PA i PA inn n= + + + + + + +− −( ) ( ) ( )1 1 12 1L (2.22)

where FAn is the future amount at the end of the nth year.Equation (2.22) is a geometric series that can be summed as follows:

Multiply both sides of Equation (2.22) by (1 + i) to obtain

( ) ( ) ( ) ( ) (1 1 1 1 12 2+ = + + + + + + +−i FA PA i PA i PA i PAnnL ++ i n) (2.23)

By subtracting Equation (2.22) from Equation (2.23), we get

( ) ( )1 1+ − = + −i FA FA PA i PAn nn (2.24)

After rearranging Equation (2.24), we obtain

FA

PA iin

n

= + −[( ) ]1 1 (2.25)

Example 2.5Assume that a person deposits $30,000 at the end of each year for the next 8 years. Calculate the total future amount of the money deposited after the 8-year period, if the annual compound interest rate is 5%.


FA = + −

=

( , )( . )

.

$ , .

30 0001 0 05 1

0 05

286 473 26

8

Thus, the total future amount of the money deposited after the 8-year period is $286,473.26.

K10869_Book.indb 17 8/26/09 1:47:15 PM


2.5.4 Uniform Periodic Payment Present Value Formula

This formula is concerned with determining the present value or worth at the end of n interest periods (years) of equal payments made at the end of each inter-est period. All payments are invested at an annual compound interest rate i.

The formula is developed as follows: At the end of the first year, after the first payment, the present value of that payment from Equation (2.14) is

V

PAip1 1

=+( ) (2.26)

whereVp1 is present value of the payment, PA, made at the end of the first year.i is annual compound interest rate.

At the end of the second year, after the second payment, the present value of that payment from Equation (2.14) is

V

PAip2 21

=+( ) (2.27)

where Vp2 is present value of the payment, PA, made at the end of the sec-ond year.

Similarly, at the end of the nth year, after the nth payment, the present value of that payment from Equation (2.14) is

V

PAipn n

=+( )1 (2.28)

whereVpn is present value of the payment, PA, made at the end of the nth year.n is number of interest periods or years.

Using Equations (2.26)–(2.28), we get the following equation for the present value of all payments:

PV V V V

PAi

PAi

PAi

p p pn= + + +

=+

++

+ ++

1 2

21 1 1

L

L( ) ( ) ( )nn

(2.29)

Equation (2.29) is a geometric series that can be summed as follows: Multiply both sides of Equation (2.29) by 1

1( )+i to obtain

PAi

PAi

PAi

PAi n( ) ( ) ( ) ( )1 1 1 12 3 1+

=+

++

+ ++ +

L (2.30)

K10869_Book.indb 18 8/26/09 1:47:50 PM


By subtracting Equation (2.29) from Equation (2.30), we obtain

PVi

PVPA

iPA

in( ) ( ) ( )1 1 11+− =

+−

++ (2.31)

After rearranging Equation (2.31), we get

PV PA

ii

n

= − +

−1 1( ) (2.32)

Example 2.6Assume that a person deposits $50,000 at the end of each year for the next 5 years. Calculate the present value of all payments, if the annual compound inter-est rate is 4%.


PV = − +

=

−( , )

( . ).

$ , .

50 0001 1 0 04

0 04

222 591 1

5

Thus, the present value of all payments is $222,591.10.

2.5.5 Formulas to Calculate Value of Annuity Payments When Annuity’s Present and Future Values Are Given

An annuity is a series of equal payments at equal time intervals. Thus, from Equation (2.25) the value of annuity payments when the future value of the annuity is known is given by

PA

FA iifn

nυ =+ −

( )( )( )1 1 (2.33)

wherePAfu is the value of annuity payments when the future value of the annuity

is given.FAn is the future value of the annuity after n interest periods or years. n is total number of interest periods or years.i is annual compound interest rate.

Similarly, from Equation (2.32), the value of annuity payments when the present value of the annuity is given is expressed by

PA

PV iip nυ =

− + −

( )( )( )1 1 (2.34)

K10869_Book.indb 19 8/26/09 1:48:18 PM


wherePApu is the value of annuity payments when the present value of the annu-

ity is known.PV is present value of all payments.

Example 2.7Assume that a firm plans to acquire a facility at the end of the next 5 years. The estimated cost of the facility after the specified period is $800,000. The firm has decided to make deposits of equal amounts of money at the end of each of next 5 years so that the total amount accumulates to $800,000. Calculate the amount of money the firm should deposit at the end of each year, if the annual compound interest rate is 8%.


PAfυ =+ −

=

( , ) ( . )( . )

$ , .

800 000 0 081 0 08 1

136 365 16

5

This means that the firm should deposit $136,365.16 at the end of each year to fulfill its objective.

Example 2.8Assume that we have the following data values:

PV = $400,000, i = 4%, and n = 7 years

Using Equation (2.34), calculate the value of annuity payments.Using the given data values in Equation (2.34) yields

PApυ =− +

=

−

( , ) ( . )( . )

$ , .

400 000 0 041 1 0 04

66 643 84

7

Thus, the value of annuity payments is $66,643.84.

2.6 Depreciation Methods

The term depreciation simply means decline in value. There are differ-ent types of depreciation with respect to engineering equipment: mon-etary depreciation, technological depreciation, physical depreciation, and

K10869_Book.indb 20 8/26/09 1:48:28 PM


functional depreciation [3]. Over the years, a number of methods with respect to monetary depreciation have been developed. Three of these methods are presented next [2–4].

2.6.1 Sum-of-Years-Digits (SYD) Method

The name of this method is derived from the calculation procedure used. The method provides a larger depreciation charge during early life years of the equipment, system, or product than during its later life years.

The annual depreciation charge is expressed by [2,4]

DC C VL n

L

C V

a a sS

S

a

= −− +

+ + + +

= −

( )( )

( )

(

11 2 3 L

SS S S SL n L L)( )( )/ ( )2 1 1− + + (2.35)

whereDCa is annual depreciation charge.Ca is product or item acquisition cost.VS is product or item salvage value at the end of its service life.LS is product or item service life expressed in years.n is total number of years of the product or item in actual service.

The book value of the product or item at the end of year n is given by [4]

V C V

L nL L

Vbn a sS

S SS= −

+ + + + −+

+2

1 2 31

( )( )

( )L

(2.36)

where Vbn is product or item book value at the end of year n.

Example 2.9Assume that the cost, useful life, and salvage value after the useful life of an engi-neering system are $900,000, 10 years, and $60,000, respectively. Calculate the system book value at the end of year 5 by using the SYD method.

By substituting the given data values into Equation (2.36), we obtain

Vb5 2 900 000 60 0001 2 3 10 5

10 10 1= − + + + + −

+( , , )

( )( )L

+

=

60 000

289 090 9

,

$ , .

Thus, the system book value at the end of year 5 is $289,090.90.

K10869_Book.indb 21 8/26/09 1:48:50 PM


2.6.2 Straight-Line Method

This method assumes the linear decrease with time in the value of an item, product, or system. Thus, during the service life of the item, product, or system an equal sum of money is charged each year for depreciation. The annual depreciation is expressed by

DC C V La a S S= −( )/ (2.37)

The book value of the product, item, or system at the end of year n is given by

V C n DCbn a a= − ( ) (2.38)

Using Equation (2.37) in Equation (2.38) yields

V C n

C VLbn a

a S

S

= − −

( ) (2.39)

Example 2.10Assume that the acquisition cost, the expected useful life, and salvage value after the useful life of a piece of equipment are $600,000, 12 years, and $30,000, respectively. The equipment annual depreciation charge is constant. Calculate the equipment annual depreciation charge.


DCa = −

=

( , , )/

$ ,

600 000 30 000 12

47 500

Thus, the equipment annual depreciation charge is $47,500.

2.6.3 Declining-Balance Method

This method is also known as the Matheson formula or the constant per-centage method. In this approach, the annual depreciation is a fixed per-centage of the book value at the beginning of the year. Although the annual depreciation is different for each year, the declining-balance (i.e., fixed-per-centage) factor remains constant throughout the useful life of the equipment or item.

This method writes off the cost of the equipment or item early in its life at an accelerated rate and at correspondingly lower annual charges close to the final years of the equipment or item service. The depreciation factor or rate

K10869_Book.indb 22 8/26/09 1:49:13 PM


is expressed by

R

VCd

S

a

LS

= −

1

1/

(2.40)

where Rd is the depreciation rate or factor. Note that this method assumes that the salvage value of the equipment or item is always positive.

The book value of the equipment or item at the end of year n is defined by

V C Rbn a dn= −( )1 (2.41)

By inserting Equation (2.40) into Equation (2.41), we get

V C

VCbn a

S

a

n LS

=

/

(2.42)

The annual depreciation charge is defined by

DC V Ra b n d= −( )[ ][ ]1 (2.43)

where Vb(n–1) is the equipment or item book value at (n – 1) years.Using Equation (2.40) in Equation (2.43) yields

DC VVCa b n

S

a

LS

= −

−[ ]( )

/

1

1

1 (2.44)

Example 2.11Assume that the cost, useful life, and salvage value after the useful life of a piece of engineering equipment are $700,000, 15 years, and $80,000, respectively. Calculate the equipment book value at the end of year 10 by using the declining-balance method.

By substituting the specified data values into Equation (2.42), we obtain

Vb10

10 15

700 00080 000700 000

164 851

=

=

( , ),,

$ ,

/

..4

Thus, the equipment book value at the end of year 10 is $164,851.40.

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Problems

1. What is the difference between simple interest and compound interest?

2. Define the following terms:present value•future amount•depreciation•

3. A company borrowed $400,000 for a period of 5 years at an annual simple interest rate of 6% to procure an engineering system. Calculate the total amount of money the company has to pay to the lender at the end of 5 years.

4. Prove the following equation:

A P inn= +( )1 (2.45)

where n is the number of interest periods.

An is the total or compound amount at the end of the nth interest period.

P is the principal amount (i.e., borrowed).

5. What is effective annual interest rate? 6. An individual deposited $90,000 in a bank at a nominal interest

rate of 7% compounded monthly, for 10 months. Estimate the value of the effective annual interest rate.

7. Assume that the total operation and maintenance cost of an engi-neering system after its 7-year usage will be $100,000. Calculate the present value of $100,000 if the annual compound interest rate is 4%.

8. A company plans to procure a facility at the end of the next 7 years. The estimated cost of the facility after the specified period is $1,000,000. The company has decided to make deposits of equal sums of money at the end of each of the next 7 years so that the total amount accumulates to $1,000,000. Calculate the amount of money the company should deposit at the end of each year, if the annual compound interest is 6%.

9. Assume that the cost, useful life, and salvage value after the useful life of a piece of engineering equipment are $660,000, 8 years, and $40,000, respectively. The equipment annual depreciation charge is constant. Calculate the equipment annual depreciation charge by using the straight-line method.

10. Compare the SYD and declining-balance depreciation methods.

K10869_Book.indb 24 8/26/09 1:49:54 PM


References

1. Paul-DeGarmo, E., Canada, J. R., and Sullivan, W. G. 1979. Engineering economy. New York: Macmillan Publishing Company, Inc.


3. Riggs, J. L. 1981. Production systems: Planning, analysis, and control. New York: John Wiley & Sons.

4. Riggs, J. L. 1968. Economic decision models for engineers and managers. New York: McGraw–Hill Book Company.

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K10869_Book.indb 26 8/26/09 1:49:54 PM

27

3Life Cycle Costing Fundamentals

3.1 Introduction

Past experience indicates that engineering equipment procured at the lowest cost may not necessarily be that which also costs the least amount of money over its useful life. More specifically, the equipment ownership cost could be quite significant and frequently exceeds the procurement cost. For example, various studies performed by the U.S. Department of Defense indicate that the maintenance cost over equipment’s useful life could be many times the procurement cost [1,2].

In fact, by simply examining the Defense Department’s overall annual budget, it can easily be observed that operation and maintenance costs are an important factor. For example, in fiscal year 1974, 27% of the overall budget of the Department of Defense accounted for operation and maintenance activi-ties and 20% was for procurement [3,4]. This simply means that, in equip-ment acquisition analysis, it is important to consider the cost of equipment ownership. Otherwise, procurement decisions based totally on the acquisi-tion cost may not be the best decision in the long term.

The approach used for estimating the total life cycle cost of equipment procurement is known as life cycle costing. This chapter presents various fundamental aspects of this approach.

3.2 Need and Information Required for Life Cycle Costing

Life cycle costing is increasingly being used in the industrial sector around the world to make various types of decisions that directly or indirectly con-cern engineering equipment and systems. There could be many reasons for this upward trend, such as [4]

competition;•increasing operation and maintenance costs;•budget limitations;•expensive products or systems (e.g., military systems, space systems, •and aircraft);

K10869_Book.indb 27 8/26/09 1:49:54 PM


rising inflation; and•increasing awareness of cost effectiveness among product, equip-•ment, and system users.

Various types of information are required to perform life cycle costing studies. These include the acquisition cost of the item, the useful operational life of the item in years, the annual maintenance cost of the item, transpor-tation (delivery) and installation costs of the item, discount and escalation rates, the annual operating cost of the item, taxes (e.g., tax benefits from depreciation, investment tax credit), and the salvage value or disposal cost of the item [5].

In any case, prior to starting a life cycle costing study, it is considered use-ful to seek answers to questions on topics such as the following [6,7]:

goal of the estimate;•assumptions and ground rules;•treatment of uncertainties;•required data;•required details of the analysis and analysis-related constraints;•involved personnel and the responsibility of the cost analyst;•controlling and auditing the life cycle costing process by the seller’s •and purchaser’s management;estimating procedures to be followed;•life cycle cost analysis users;•life cycle cost analysis format;•life cycle costing time schedule;•required accuracy and precision of the analysis; and•fund limitations.•

3.3 Life Cycle Costing Application Areas

Life cycle costing can be used in a large number of areas. The six primary uses of life cycle cost include [6]:

selecting among competing bidders for a project;•long-range planning and budgeting;•controlling an ongoing project;•comparing competing projects;•

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Life Cycle Costing Fundamentals 29

deciding the replacement of aging equipment; and•comparing logistics concepts.•

Lamar [8] has presented the following somewhat more specific applica-tions of life cycle cost analyses:

determining cost drivers;•forecasting future budget needs;•selecting the most effective procurement strategy;•improving comprehension of fundamental design-related param-•eters in equipment or system product design and development;formulating contractor incentives;•making strategic decisions and design trade-offs;•optimizing appropriate training needs;•choosing among options;•providing effective objectives for program control;•assessing new technology application; and•carrying out source selections.•

3.4 Life Cycle Costing Activities and Associated Steps

Many activities are associated with life cycle costing. Some of these include [9]:

defining an item’s or a product’s life cycle;•identifying all cost drivers;•establishing escalated and discounted life cycle costs;•developing an accounting breakdown structure;•establishing cost estimating relationships for each and every compo-•nent in the life cycle cost breakdown structure;developing constant dollar cost profiles;•defining activities that generate an item’s or a product’s owner-•ship costs;conducting appropriate sensitivity analysis; and•identifying cause and effect relationships.•

Over the years, many authors have proposed steps for performing life cycle cost analysis [10–13]. Figure 3.1 shows 10 steps considered quite effective in

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performing life cycle cost analysis [14]. Additional information on these steps is available in reference 14.

3.5 Approach for Incorporating Life Cycle Costing into the Planning Process for Proposals and Contracts

Over the years, equipment or system procurement contracts requiring con-tractor or manufacturer commitments for equipment or system life cycle cost have increased quite significantly. Many of these contractors and

Determine life cycle cost analysis objective

Define and scope the system/support system

Choose the effective estimating methodology/life cycle cost models

Obtain all essential data and make theappropriate inputs to the selected methodology/model

Conduct sanity checks of outputs and inputs

Conduct essential sensitivity analysis and riskassessment

Formulate life cycle cost analysis results

Document the life cycle cost analysis

Present the life cycle cost analysis asappropriate

Update the life cycle cost analysis asappropriate

FIGURE 3.1Steps for performing life cycle cost analysis.

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manufacturers are not familiar with life cycle cost-related acquisitions. In order to overcome this shortcoming, a six-step approach for these contrac-tors and manufacturers to prepare for life cycle cost-related acquisitions follows [15]:

Organize for life cycle costing.• This step is basically concerned with establishing a proper organization for life cycle costing and assign-ing life cycle cost responsibilities.Gather and develop background information related to life cycle costing.• This step calls for becoming acquainted with the existing life cycle cost estimation models and components of the life cycle cost consid-ered vital to the company’s product and equipment.Perform analysis of all requirements for life cycle costing–related response.• This step involves tasks such as performing analysis of likely life cycle cost estimation model components to determine the types of data required for life cycle cost response and performing analysis of the information considered essential for management decision making.Develop a plan for the life cycle costing technical proposal.• This step is basically concerned with planning the life cycle costing–related response for a technical proposal under consideration.Develop a plan to identify and analyze life cycle cost risk. • This step calls for developing a plan to identify risk areas and address methods to analyze such risks when life cycle cost–related guarantees are com-mitted as an element of a proposed procurement.Develop a plan to achieve life cycle cost goals.• This step involves develop-ing a plan to achieve the set life cycle cost goals during the specified contract period.

3.6 Areas for Evaluating a Life Cycle Costing Program

In order to keep a life cycle costing program in good order, it is essential to evaluate it periodically. There are many areas in which questions could be raised to determine the effectiveness of the life cycle costing program. Some of these areas include [4,6,16]:

effectiveness of cost-estimating techniques used;•cost model construction;•broadness of cost-estimating database;•identification of all cost drivers;•proper consideration of discounting and inflation factors;•

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performance of trade-off studies;•inclusion of all life cycle costing–related requirements into design •subcontracts;cost performance review of subcontractors;•cost estimates’ validation through an independent appraisal;•life cycle costing management representative’s qualifications;•coordination of life cycle cost and design to cost-related activities;•defining of cost priority with respect to factors such as product perfor-•mance, delivery schedule, and other requirements by management;formal notifications to all organizations or departments involved in •the life cycle costing program regarding their cost goals;compatibility of system safety, reliability, and maintainability pro-•grams with life cycle cost–related requirements; andawareness of the buyer regarding the top 10 cost drivers and proper •suggestions to reduce such costs.

3.7 Life Cycle Costing Data Sources

In order to perform effective life cycle cost analysis, the availability of reli-able cost data is vital. This means that the existence of good cost data banks is very important. Thus, in developing a new cost data bank, careful atten-tion must be given to factors such as comprehensiveness, size, uniformity, flexibility, responsiveness, ready accessibility, orientation, and expansion or contraction capability [17]. Furthermore, at a minimum, a life cycle cost-ing data bank should incorporate information such as user pattern records, descriptive records (hardware and site), cost records, and procedural records (operation and maintenance).

Although data for life cycle cost analysis can be obtained from many sources, their amount and quality may vary quite considerably. Therefore, prior to starting a life cycle cost study, it is important to examine carefully factors such as data bias, data applicability, data availability, data compa-rability to other existing data, data orientation toward the problem under consideration, and data coordination with other information. Some of the important sources for obtaining cost-related data include [4,17,18]:

costs for pressure vessels [19];•American Building Owners and Managers Association (BOMA) •handbook;costs for solid waste shredders [20];•

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costs for heat exchangers [21–23];•unit price manuals: Marshall and Swift, means, Dodge, Richardson, •and building cost file;cost analysis cost estimating (CACE) model [24,25];•costs for varieties of process equipment [26–29];•budgeting annual cost estimating (BACE) model [24,25];•programmed review of information for costing and evaluation •(PRICE) model [24]; andcosts for motors, storage tanks, centrifugal pumps, etc. [30,31].•

3.8 Life Cycle Costing Advantages and Disadvantages and Related Important Points

Over the years, various advantages and disadvantages of life cycle costing have been identified by various professionals. Some of the important advan-tages of life cycle costing are shown in Figure 3.2 [4]. In contrast, some of the main disadvantages of life cycle costing include that it

is time consuming;•is costly;•has doubtful data accuracy; and•is a trying task when attempting to obtain data for analysis.•

Useful to reduce the total cost

A useful tool for making decisions associated with equipment replacement, planning, and budgeting

Useful to control programs

Useful in comparing the cost of competing projects

An excellent tool for making a selection among the competing contractors/manufacturers

Advantages

FIGURE 3.2Life cycle costing advantages.

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Many important points are associated with life cycle costing, some of which include:

The main goal of life cycle costing is to get the maximum benefit •from limited resources.Management plays a key role in making life cycle costing a worth-•while effort.Risk management is the essence of life cycle costing in general.•The availability of good data is very important for good life cycle •cost estimates.The life cycle cost model must include all program-related costs.•There is a definite need for both the product manufacturer and the •user to organize effectively to control life cycle cost.There is a definite need to perform trade-offs among life cycle cost, •design to cost, and performance throughout the life of the program.Some surprises may still occur, even when the estimator is very •competent.Life cycle costing is gaining importance as a method for perform-•ing design optimization, making strategic decisions, conducting detailed trade-off studies, etc.A highly knowledgeable and experienced cost analyst may compen-•sate for various database-related difficulties.

3.9 Life Cycle Costing Concept Application in Selecting Equipment from Competing Manufacturers

From time to time, equipment or system users are faced with selecting the most cost-effective equipment or system from a number of competing manu-facturers. In situations such as these, life cycle costing becomes a useful tool. The application of the life cycle costing concept in selecting the most cost-ef-fective equipment from competing manufacturers is demonstrated through Example 3.1.

Example 3.1A company using machining equipment to manufacture a certain type of engi-neering part is contemplating replacing it with a better version. Four different pieces of machining equipment, manufactured by four different manufacturers, are being considered for its replacement; their data are presented in Table 3.1.

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Determine which of the four pieces of machining equipment should be procured to replace the existing one in regard to their life cycle costs.

Life Cycle Cost Analysis: Machining Equipment A

The expected cost, Cfa, of failure per year of machining equipment A is given by

Cfa =

=

( , ) ( . )

$

2 000 0 08

160

where Cfa is the machining equipment A annual expected failure cost.Using Chapter 2 and reference 4, the present value, PVaf, of machining equip-

ment A life cycle failure cost is expressed by

PV C

iiaf fa

k

= − +

−1 1( ) (3.1)

wherePVaf is present value of machining equipment A life cycle failure cost.i is annual interest rate.k is machining equipment’s expected useful life in years.

By substituting the preceding calculated value and the given data values into Equation (3.1), we get

PVaf = − +

=

−( )

( . ).

$ .

1601 1 0 06

0 06

1176 61

10

TABLE 3.1

Data for Four Types of Machining Equipment under Consideration

No. Description

Machining Equipment

A

Machining Equipment

B

Machining Equipment

C

Machining Equipment

D

1 Procurement cost $300,000 $270,000 $290,000 $350,0002 Expected useful life in years 10 10 10 103 Annual failure rate 0.08 0.07 0.06 0.044 Cost of a failure $2,000 $2,500 $3,000 $1,0005 Annual interest rate 6% 6% 6% 6%6 Annual operating cost $6,000 $7,000 $6,500 $8,000

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Similarly, using Chapter 2 and reference 4, the present value, PVao , of machining equipment A life cycle operating cost is given by

PV C

iiao oa

k

= − +

−1 1( ) (3.2)

wherePVao is present value of machining equipment A life cycle operating cost.Coa is machining equipment A annual operating cost.


PVao = − +

=

−( , )

( . ).

$ , .

6 0001 1 0 06

0 06

44 160 52

10

Thus, the life cycle cost of machining equipment A is given by

LCC PC PV PVa a af ao= + + (3.3)

whereLCCa is machining equipment A life cycle cost.PCa is machining equipment A procurement cost.

By substituting the given data value and the preceding calculated values into Equation (3.3), we obtain

LCCa = + +

=

300 000 1176 61 44 160 52

345 337 13

, . , .

$ , .

Life Cycle Cost Analysis: Machining Equipment B

The expected cost, Cfb, of failure per year of machining equipment B is given by

Cfb =

=

( , )( . )

$

2 500 0 07

175

where Cfb is machining equipment B annual expected failure cost.Using Chapter 2 and reference 4, the present value, PVbf, of machining equip-

ment B life cycle failure cost is given by

PV C

iibf fb

k

= − +

−1 1( ) (3.4)

where PVbf is present value of machining equipment B life cycle failure cost.

K10869_Book.indb 36 8/26/09 1:50:39 PM


By substituting the preceding calculated value and the given data values into Equation (3.4), we get

PVbf = − +

=

−( )

( . ).

$ , .

1751 1 0 06

0 06

1 288 01

10

Similarly, using Chapter 2 and reference 4, the present value, PVbo, of machining equipment B life cycle operating cost is expressed by

PV C

iibo ob

k

= − +

−1 1( ) (3.5)

wherePVbo is present value of machining equipment B life cycle operating cost.Cob is machining equipment B annual operating cost.


PVbo = − +

=

−( , )

( . ).

$ , .

7 0001 1 0 06

0 06

51 520 61

10

Thus, the life cycle cost of machining equipment B is given by

LCC PC PV PVb b bf bo= + + (3.6)

whereLCCb is machining equipment B life cycle cost.PCb is machining equipment B procurement cost.

By substituting the given data value and the preceding calculated values into Equation (3.6), we get

LCCb = + +

=

270 000 1 288 01 51 520 61

322 808 62

, , . , .

$ , .

Life Cycle Cost Analysis: Machining Equipment C

The expected cost, Cfc, of failure per year of machining equipment C is given by

Cfc =

=

( , )( . )

$

3 000 0 06

180

where Cfc is machining equipment C annual expected failure cost.

K10869_Book.indb 37 8/26/09 1:51:07 PM


Using Chapter 2 and reference 4, the present value, PVcf, of machining equip-ment C life cycle failure cost is expressed by

PV C

iicf fc

k

= − +

−1 1( ) (3.7)

where PVcf is present value of machining equipment C life cycle failure cost.By substituting the preceding calculated value and the given data values into

Equation (3.7), we get

PVcf = − +

=

−( )

( . ).

$ , .

1801 1 0 06

0 06

1 324 81

10

Similarly, using Chapter 2 and reference 4, the present value, PVco, of machining equipment C life cycle operating cost is expressed by

PV C

iico oc

k

= − +

−1 1( ) (3.8)

wherePVco is present value of machining equipment C life cycle operating cost.Coc is machining equipment C annual operating cost.


PVco = − +

=

−( , )

( . ).

$ , .

6 5001 1 0 06

0 06

47 840 56

10

Thus, the life cycle cost of machining equipment C is given by

LCC PC PV PVc c cf co= + + (3.9)

whereLCCc is machining equipment C life cycle cost.PCc is machining equipment C procurement cost.


LCCc = + +

=

290 000 1 324 81 47 840 56

339 165 37

, , . , .

$ , .

K10869_Book.indb 38 8/26/09 1:51:34 PM


Life Cycle Cost Analysis: Machining Equipment D

The expected cost, Cfd, of failure per year of machining equipment D is given by

Cfd =

=

( , )( . )

$

1 000 0 04

40

where Cfd is machining equipment D annual expected failure cost.Using Chapter 2 and reference 4, the present value, PVdf, of machining equip-

ment D life cycle failure cost is expressed by

PV C

iidf fd

k

= − +

−1 1( ) (3.10)

where PVdf is present value of machining equipment D life cycle failure cost.By substituting the preceding calculated value and the given data values into

Equation (3.10), we get

PVdf = − +

=

−( )

( . ).

$ .

401 1 0 06

0 06

294 40

10

Similarly, using Chapter 2 and reference 4, the present value, PVdo, of machining equipment D life cycle operating cost is given by

PV C

iido od

k

= − +

−1 1( ) (3.11)

wherePVdo is present value of machining equipment D life cycle operating cost.Cod is machining equipment D annual operating cost.


PVdo = − +

=

−( , )

( . ).

$ , .

8 0001 1 0 06

0 06

58 880 69

10

Thus, the life cycle cost of machining equipment D is expressed by

LCC PC PV PVd d df do= + + (3.12)

whereLCCd is machining equipment D life cycle cost.PCd is machining equipment D procurement cost.

K10869_Book.indb 39 8/26/09 1:52:02 PM



LCCd = + +

=

350 000 294 40 58 880 69

409 175 09

, . , .

$ , .

Thus, the life cycle costs of machining equipment A, B, C, and D are $345,337.13, $322,808.62, $339,165.37, and $409,175.09, respectively. By examining these values, it is concluded that machining equipment B should be purchased because its life cycle cost is the lowest.

Problems

1. Write an essay on life cycle costing fundamentals. 2. Discuss the need for life cycle costing. 3. List at least 10 specific applications of life cycle cost analyses. 4. List at least eight activities associated with life cycle costing. 5. What are the steps used to perform life cycle cost analysis? 6. Describe the six-step approach for unfamiliar contractors and

manufacturers to prepare for life cycle cost–related acquisitions. 7. List at least 12 areas on which questions could be raised to deter-

mine the effectiveness of a life cycle costing program. 8. List at least 10 important sources for obtaining cost-related data. 9. What are the advantages and disadvantages of life cycle costing? 10. A company using a machine to manufacture a certain type of

engineering part is contemplating replacing it with a better one. Two different machines are being considered for its replacement and their data are presented in Table 3.2. Determine which of the two machines should be procured to replace the existing machine in regard to their life cycle costs.

TABLE 3.2

Data for Two Machines under Consideration

No. Description Machine A Machine B

1 Procurement cost $140,000 $170,0002 Annual failure rate 0.03 0.043 Expected useful life in years 12 124 Annual operating cost $6,000 $4,0005 Cost of a failure $12,000 $13,0006 Annual interest rate 8% 8%

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References


2. Dhillon, B. S. 1983. Reliability engineering in systems design and operations. New York: Van Nostrand Reinhold Company.

3. Louis-Wienecke, E., and Feltus, E. E. 1979. Predictive operations and mainte-nance cost model. Report no. ADA078052. Available from the National Technical Information Service (NTIS), Springfield, VA.


5. Brown, R. J. 1979. A new marketing tool: Life cycle costing. Industrial Marketing Management 8:109–113.

6. Robert-Seldon, M. 1979. Life cycle costing: A better method of government procure-ment. Boulder, CO: Westview Press.

7. Reiche, H. 1980. Life cycle cost. In Reliability and maintainability of electronic sys-tems, ed. J. E. Arsenault and J. A. Roberts, 3–23. Potomac, MD: Computer Science Press.

8. Lamar, W. E. 1981. Technical evaluation report on design to cost and life cycle cost. North Atlantic Treaty Organization Advisory Group for Aerospace Research and Development (AGARD) advisory report no. 165. Available from the National Technical Information Service (NTIS), Springfield, VA.

9. Earles, M. 1981. Factors, formulas, and structures for life cycle costing. Concord, MA: Eddins–Earles.

10. Kaufman, R. J. 1969. Life cycle costing: Decision making tool for capital equip-ment acquisitions. Journal of Purchasing 5:16–31.

11. Kaufman, R. J. 1969. Life cycle costing: For capital equipment decisions. Automation March: 75–80.

12. Coe, C. K. 1981. Life cycle costing by state governments. Public Administration Review September/October: 564–569.

13. Wynholds, H. W., and Skratt, J. P. 1977. Weapon system parametric life cycle cost analysis. Proceedings of the Annual Reliability and Maintainability Symposium 303–309.

14. Greene, L. E., and Shaw, B. L. 1990. The steps for successful life cycle cost analy-sis. Proceedings of the IEEE National Aerospace and Electronics Conference 1209–1216.

15. Schmidt, B. A. 1979. Preparation for LCC proposals and contracts. Proceedings of the Annual Reliability and Maintainability Symposium 62–66.

16. Bidwell, R. L. 1977. Checklist for evaluating LCC program effectiveness. Product Engineering Services Office, U.S. Department of Defense, Washington, D.C.

17. Bowen, B., and Williams, J. 1975. Life costing and problems of data. Industrialization Forum 6:21–24.

18. Dhillon, B. S. 1999. Design reliability: Fundamentals and applications. Boca Raton, FL: CRC Press.

19. Mulet, A., Corripio, A. B., and Evans, L. B. 1981. Estimate cost of pressure ves-sels via correlations. Chemical Engineering 88:20, 456.

20. Fang, C. S. 1980. The cost of shredding municipal solid waste. Chemical Engineering 87 (7): 151–153.

K10869_Book.indb 41 8/26/09 1:52:07 PM


21. Purohit, G. P. 1985. Cost of double pipe and multi-tube heat exchangers. Chemical Engineering 92:96–97.

22. Woods, D. R., Anderson, S. J., and Norman, S. L. 1976. Evaluation of capital cost data: Heat exchangers. Canadian Journal of Chemical Engineering 54:469.

23. Kumana, J. D. 1984. Cost update on specialty heat exchangers. Chemical Engineering 91 (13): 164.

24. Marks, K. E., Garrison-Massey, H., and Bradley, B. D. 1978. An appraisal of mod-els used in life cycle cost-estimation for U.S. Air Force (USAF) aircraft systems. Report no. R-2287-AF. Prepared by the Rand Corporation, Santa Monica, CA.

25. Department of the Air Force. 1975. USAF cost and planning factors. Report no. AFR 173-10, Washington, D.C.

26. Hall, R. S., Mately, J., and McNaughton, K. J. 1982. Current costs of process equipment. Chemical Engineering 87 (7): 80.

27. Klumpar, I. V., and Slavsky, S. T. 1985. Updated cost factors: Process equip-ment, commodity materials, and installation labor. Chemical Engineering 92 (15): 73–74.

28. Humphreys, K. K., and Katell, S. 1981. Basic cost engineering. New York: Marcel Dekker.

29. Peters, M. S., and Timmerhaus, K. D. 1980. Plant design and economics for chemical engineers. New York: McGraw–Hill Book Company.

30. Corripio, A. B., Chrien, K. S., and Evans, L. B. 1982. Estimate costs of heat exchangers and storage tanks via correlations. Chemical Engineering 89 (2): 125–126.

31. Corripio, A. B., Chrien, K. S., and Evans, L. B. 1982. Estimate costs of centrifugal pumps and electric motors. Chemical Engineering 89 (4): 115.

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43

4Life Cycle Cost Models and Cost Estimation Methods

4.1 Introduction

Over the years, a large number of life cycle cost models have been devel-oped that include both general and specific models [1,2]. No single life cycle cost model has been accepted as a standard model in the industrial sector. There could be many reasons for not having a standard model, including the inclinations of users, the nature of the problem, the existence of many dif-ferent cost data collection systems, and many different types of equipment, devices, or systems. Nonetheless, irrespective of the types of models used in performing life cycle cost analysis, they all must be effective in representing equipment, systems, or subsystems, transparent and visible.

Cost estimating is an important activity because estimated cost has to be as close as possible to actual value; otherwise, an incorrect estimate may lead to serious consequences of various types. Currently, many methods are used to estimate various types of costs. Each one has its advantages and disadvan-tages. More specifically, a cost estimation method or approach may be very effective in one type of application and rather weak in another. This simply means that utmost care is necessary in selecting a cost estimation method for a specific application.

This chapter presents some of the life cycle cost models and cost estima-tion methods considered useful in performing life cycle cost analysis.

4.2 Types of Life Cycle Cost Models and Their Inputs

Over the years, life cycle cost models have been classified under various cate-gories [3–6]. For example, Gupta [3] and Sherif and Kolarik [5] have classified life cycle cost models under three categories: conceptual models, analytical models, and heuristic models. The conceptual models are quite flexible but have rather limited applications; they are usually based on the hypothesized relationships of variables given in a qualitative fashion. One example of the conceptual models is available in Goldman and Slattery [7].

K10869_Book.indb 43 8/26/09 1:52:07 PM


The analytical models are based on some sort of mathematical relationship and their subcategories include logistic support models, design trade mod-els, and the total cost models. Finally, the heuristic models may be described simply as the ill-structured version of the analytical models. An example of these models is available in Kolarik [8]. Overall, in this chapter, the life cycle cost models are simply classified under two categories: general life cycle cost models and specific life cycle cost models.

There are many inputs to life cycle cost models. These include [6,9]:

warranty coverage period;•average material cost of a failure;•cost of training;•cost of installation;•system’s or item’s listed price;•cost of carrying spares in inventory;•mean time between failures;•mean time to repair;•spares’ requirements;•cost of labor per corrective maintenance action; and•time spent for travel.•

4.3 General Life Cycle Cost Models

The general life cycle cost models are not tied to any specific system or equip-ment. Some of these models are presented next.

4.3.1 General Life Cycle Cost Model I

In this case, the equipment or system life cycle cost is divided into two main parts: recurring cost and nonrecurring cost. Thus, the system or equipment life cycle cost is expressed by [10]

LCC RC NRC= + (4.1)

whereLCC is item or system life cycle cost.RC is recurring cost.NRC is nonrecurring cost.

K10869_Book.indb 44 8/26/09 1:52:10 PM

Life Cycle Cost Models and Cost Estimation Methods 45

The recurring cost, RC, is expressed by

RC OC IC SC MC MTC= + + + + (4.2)

whereOC is operating cost.IC is inventory cost.SC is support cost.MC is manpower cost.MTC is maintenance cost.

The nonrecurring cost, NRC, is expressed by

NRC C C C C C C Cp i q r t rm s= + + + + + + (4.3)

whereCp is procurement cost.Ci is installation cost.Cq is qualification approval cost.Cr is research and development cost.Ct is training cost.Crm is reliability and maintainability improvement cost.Cs is support cost.

4.3.2 General Life Cycle Cost Model II

In this case, the equipment or system life cycle cost is divided into three main parts: procurement cost, initial logistic cost, and recurring cost. Thus, the system or equipment life cycle cost is expressed by [11]

LCC C C C= + +1 2 3 (4.4)

whereLCC is item or system life cycle cost.C1 is acquisition or procurement cost.C2 is initial logistic cost.C3 is recurring cost.

The initial logistic cost, C2, is composed of one-time costs such as the cost of procurement of new support equipment not accounted for in the life cycle costing solicitation and training, the cost of existing support equipment modifications, and the cost of initial technical data management.

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The three main components of the recurring cost, C3, are operating cost, management cost, and maintenance cost.

4.3.3 General Life Cycle Cost Model III

This model was developed by the U.S. Navy to estimate life cycle cost of major weapon systems [12,13]. The system life cycle cost is divided into five main parts: research and development cost, the cost of associated systems, investment cost, termination cost, and operating and support cost. Thus, the system life cycle cost is expressed by

LCC C C C C C= + + + +1 2 3 4 5 (4.5)

whereLCC is system life cycle cost.C1 is research and development cost.C2 is cost of associated systems.C3 is investment cost.C4 is termination cost.C5 is operating and support cost.

The two main components of the research and development cost, C1, are full-scale development cost and validation cost. Similarly, the two main ele-ments of the cost of associated systems, C2, are their investment cost and their operating and support cost.

The investment cost, C3, is also made up of two main components: the gov-ernment investment cost and the procurement cost. The termination cost, C4, is expressed by

C x ci t

i

m

4

1

==∑

(4.6)

wherem is total number of years in the life cycle.xi is total number of major system items put out of action during year i.ct is terminal cost of the major system item.

Finally, the elements of the operating and support cost are depot supply cost, depot maintenance cost, operating cost, personnel support and training costs, sustaining investment cost, installation support cost, second destina-tion transportation cost, and organizational and intermediate maintenance activity cost.

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4.3.4 General Life Cycle Cost Model IV

In this case, the life cycle cost is expressed by [2,14]

LCC C C C Ccp dp pp op= + + + (4.7)

whereLCC is life cycle cost.Ccp is cost associated with the conceptual phase.Cdp is cost associated with the definition phase.Cpp is cost associated with the procurement phase.Cop is cost associated with the operational phase.

The costs of conceptual and definition phases are relatively small in compar-ison to the costs of procurement and operational phases. They are basically associated with the labor effort.

The four main elements of the procurement phase cost are the cost of the prime equipment or system, the cost of acquisition personnel, the cost of support equipment, and the cost of program management. Finally, the opera-tional phase cost is expressed by

C C C Cop m fo oa= + + (4.8)

whereCm is maintenance cost.Cfo is functional operating cost.Coa is operational administrative cost.

Additional information on this model is available in Dhillon [2] and Stordahl and Short [14].

4.3.5 General Life Cycle Cost Model V

In this case, the life cycle cost is expressed by [6,15]

LCC C C C Crd pc os rt= + + + (4.9)

whereLCC is life cycle cost.Crd is research and development cost.Cpc is production and construction cost.Cos is operation and support cost.Crt is retirement and disposal cost.

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The research and development, Crd, is expressed by

C Crd rdj

j

==

∑1

7

(4.10)

where Crdj is the jth cost element of the research and development cost forj = 1 (means product planning);j = 2 (means engineering design);j = 3 (means product or system life cycle management);j = 4 (means system or product test and evaluation);j = 5 (means product or system research);j = 6 (means product or system software); andj = 7 (means design documentation).

The production and construction cost, Cpc, is defined by

C Cpc pcj

j

==

∑1

5

(4.11)

where Cpcj is the jth cost element of the production and construction cost forj = 1 (means manufacturing);j = 2 (means construction);j = 3 (means quality control);j = 4 (means initial logistics support); andj = 5 (means industrial engineering and operations analysis).

The operation and support cost, Cos, is expressed by

C Cos osj

j

==

∑1

3

(4.12)

where Cosj is the jth cost element of the operation and support cost forj = 1 (means system or product operations);j = 2 (means product or system distribution); andj = 3 (means sustaining logistic support).

The retirement and disposal cost, Crt, is defined by

C C K C rrt ur id= + −[ ( )]θ υ (4.13)

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whereCur is ultimate retirement cost of the system or product.q is the condemnation factor.K is total number of unscheduled maintenance actions.Cid is item disposal cost.ru is reclamation value.

4.3.6 General Life Cycle Cost Model VI

This model was developed by the Material Command of the U.S. Army and is composed of three main components: investment cost, research and devel-opment cost, and operating and support cost [16–18]. Thus, the life cycle cost is expressed mathematically by [6,16–18]

LCC C C C= + +1 2 3 (4.14)

whereLCC is life cycle cost.C1 is research and development cost.C2 is investment cost.C3 is operating and support cost.

The research and development cost, C1, is composed of the following 10 components:

research and development data cost;•cost of research and development tooling;•cost of research and development facilities;•development engineering cost;•prototype manufacturing cost;•research and development test and evaluation cost;•producibility engineering and planning cost;•research and development system or project management cost;•research and development training services and equipment cost; and•other research and development costs.•

The investment cost, C2, is composed of 11 components:

cost of production;•initial training cost;•

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transportation cost;•cost of data;•cost of engineering changes;•nonrecurring investment cost;•cost of system test and evaluation;•production phase system or project management cost;•cost of initial spares and repair parts;•operational or site activation cost; and•other investment costs.•

Finally, the operating and support cost, C3, is composed of six major components:

cost of indirect support operations;•cost of depot maintenance;•cost of material modifications;•consumption cost;•cost of military personnel; and•cost of other direct support operations.•

Additional information on this model is available in references 16–18.

4.4 Specific Life Cycle Cost Models

Over the years, many mathematical models have been developed to estimate life cycle cost of specific systems or items. Some of these models are pre-sented next.

4.4.1 Specific Life Cycle Cost Model I

This model is concerned with estimating the life cycle cost of switching power supplies, which is expressed by [19]

LCC IC FCs = + (4.15)

whereLCCs is life cycle cost of switching power supplies.IC is initial cost.FC is failure cost.

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The failure cost, FC, is expressed by

FC n C Cr s= +λ ( )( ) (4.16)

wherel is unit constant failure rate.n is expected life of the product/unit.Cr is repair cost.Cs is cost of spares.

The cost of spares, Cs, is defined by

C C Ks u= ( ) (4.17)

whereCu is unit spare cost.K is fractional number of spares for each active unit.

4.4.2 Specific Life Cycle Cost Model II

This model is concerned with estimating the life cycle cost of health care facilities. The health care facility life cycle cost is expressed by [6,13]

LCC C Ch c o= + (4.18)

whereLCCh is health care facility life cycle cost.Cc is capital cost.Co is operating cost.

The capital cost, Co, is composed of the following eight cost components:

land acquisition cost;•financing cost;•collateral equipment cost;•direct construction or purchase cost;•indirect cost;•demolition and site preparation cost;•alteration and replacement cost; and•denial of use cost.•

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Similarly, the operating cost, Co, is composed of the following 19 cost components:

utilities and fuel cost;•structural maintenance cost;•heating system operations and maintenance cost;•painting cost;•equipment (furnishings) maintenance cost;•exterior building cleaning cost;•electrical system operations and maintenance cost;•space changes cost;•exterior restoration cost;•grounds and roads maintenance cost;•equipment (fixed equipment and specific construction) mainte-•nance cost;insect and rodent control cost;•incinerator and trash removal cost;•building internal cleaning cost;•special mechanical systems operations and maintenance cost;•elevator, escalator, and dumbwaiter operations cost;•plumbing and sewage systems operations and maintenance cost;•fire protection systems maintenance cost; and•air conditioning and ventilating system operations and mainte-•nance cost.

4.4.3 Specific Life Cycle Cost Model III

This model is concerned with estimating the life cycle cost of an early warn-ing radar system. The radar life cycle cost is expressed by [6]

LCC C C Cr p o s= + + (4.19)

whereLCCr is early warning radar life cycle cost.Cp is radar procurement cost.Co is radar operation cost.Cs is radar logistic support cost.

The radar procurement cost, Cp, is expressed by

C FC ICC DC DOCp = + + + (4.20)

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whereFC is fabrication cost.ICC is installation and checkout cost.DC is design cost.DOC is document cost.

The radar operation cost, Co, is defined by

C C C Co = + +1 2 3 (4.21)

whereC1 is fuel cost.C2 is cost of personnel.C3 is cost of power.

The radar logistic support cost, Cs, is expressed by

C CRL CRM CIS CRS CIT ACs = + + + + + (4.22)

whereCRL is cost of repair labor.CRM is cost of repair material.CIS is cost of initial spares.CRS is cost of replacement spares.CIT is cost of initial training.AC is age cost.

The life cycle cost predicted breakdown percentages for a specific early warn-ing radar are available in Dhillon [6].

4.4.4 Specific Life Cycle Cost Model IV

This model is concerned with estimating the life cycle cost of inertial sys-tems. The inertial systems life cycle cost is expressed by [20]

LCC RDTC P C OMCis = + + (4.23)

whereLCCis is inertial systems life cycle cost.RDTC is research, development, test, and evaluation cost.PC is procurement cost.OMC is operation and maintenance cost.

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The research, development, test, and evaluation cost, RDTC, is composed of eight elements:

software cost;•testing cost;•program management cost;•cost of conceptual studies;•cost of engineering change proposals;•cost of design engineering;•cost of technical data; and•training cost.•

The 12 distinct components of the procurement cost include:

cost of new facilities;•cost of spares;•support equipment acquisition cost;•system recurring acquisition cost;•cost of technical data;•initial training course cost;•training equipment cost;•cost of production tooling and test equipment;•production program start-up cost;•cost of initial item management;•field engineering cost; and•equipment installation cost.•

The operation and maintenance cost, OMC, is expressed by

OMC OMCji

i

n

j

===∑∑

11

3

(4.24)

wheren is total number of years.OMCji is operation and maintenance cost at the jth level of maintenance in

the ith year.

Additional information on the model is available in DeBurkarte [20].

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4.4.5 Specific Life Cycle Cost Model V

This model is concerned with estimating the life cycle cost of software. Sometimes, the model is called the “Boeing C-14 model” [21,22]. The software life cycle cost is expressed by [21,22]

LCC AC SCs s s= + (4.25)

whereLCCs is life cycle cost of software.ACs is acquisition cost of software.SCs is support cost of software.

The support cost of software, SCs, is expressed by

SC LC SMM SCs j a=

+ +∑( . )( ) ( )2 5 1 α

(4.26)

whereLC is direct labor cost per man-month.∑SMMj is required man-months for support in month j.a is overhead factor.SCa is additional (other) support costs.

Additional information on the model is available in references 21 and 22.

4.5 Cost Estimation Methods

Over the years, many methods have been developed to estimate costs [23–26]. Some of the methods considered useful for application in the area of life cycle costing are presented next.

4.5.1 Cost Estimation Method I

This method is considered quite useful to obtain quick approximate cost estimates for similar new plants, projects, or equipment of different capaci-ties. The cost-capacity relationship is defined by

C C

KKn o

n

o

=

α

(4.27)

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whereCn is cost of the new plant, project, or equipment under consideration.Co is cost of the old but similar equipment, plant, or project.Kn is capacity of the new plant, project, or equipment.Ko is capacity of old but similar equipment, plant, or project.a is the cost-capacity factor whose frequently used value is 0.6. The pro-

posed values for this factor for items such as heat exchangers, heat-ers, pumps, and tanks are 0.6, 0.8, 0.6, and 0.7, respectively [23,27,28].

Example 4.1An electric utility spent $900 million to construct a 1,000 megawatt (MW) nuclear power generating station. In order to satisfy the increasing demand for electricity, the company is planning to construct a 2,000 MW nuclear power generating sta-tion. Calculate the cost of the new station, if the value of the cost-capacity factor is 0.6.


Cn =

=

9002 0001 000

1 364 15

0 6,,

$ , .

.

million

Thus, the construction cost of the new nuclear power station will be $1,364.15 million.

4.5.2 Cost Estimation Method II

This method is known as the Lang factor method, after its originator, H. L. Lang [29]. The method is used for obtaining quick order-of-magnitude cost estimates by utilizing historical average cost factors. Lang proposed to esti-mate total plant costs from the delivered equipment cost by using three fac-tors as multipliers: n = 3.10 (for solid process plants), n = 3.63 (for solid-fluid plants), and n = 4.74 (for fluid process plants) [29].

Thus, the total estimate for plant cost is obtained by using

TPC n DEC= ( )( ) (4.28)

whereTPC is the total estimate for plant cost.n is the Lang factor, whose value depends on the nature of the plant. DEC is delivered equipment cost.

Example 4.2Assume that a fluid-processing plant’s delivered equipment cost is $40 million. Calculate the total plant cost.

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By substituting the given data value and information into Equation (4.28), we get

TPC =

=

( . ) ( )

$ .

4 74 40

189 6 million

Thus, the total plant cost will be $189.6 million.

4.5.3 Cost Estimation Method III

This method is basically a refinement of the Lang factor method and is known as the Hand method, after its originator, W. E. Hand [30]. In the refinement, Hand proposed the use of different factors for various groups of equipment.

The total installed cost for each equipment group is defined by [25,30]

IC m DECt = ( )( ) (4.29)

whereICt is total installed cost of each equipment group.m is the Hand factor that covers field materials (structures, insulation, pip-

ing, electrical, finishes, and foundations), labor, and indirect costs. The values of the Hand factor for various groups of equipment are 2 (fired heaters), 2.5 (compressors), 2.5 (miscellaneous equipment), 3.5 (heat exchangers), 4 (pumps), 4 (pressure vessels), 4 (instruments), and 4 (fractionating towers).

DEC is delivered equipment cost.

Note that the Hand factors do not incorporate a contingency allowance. Additional information on this method is available in references 25, 30, and 31.

4.5.4 Cost Estimation Method IV

This method is quite useful to make an order-of-magnitude approximation of operating labor requirements in the absence of a Manning table. The method is known as the Wessell method. Thus, the Wessell equation is expressed by [32]

OH KPλ

α=

( ) .0 76

(4.30)

whereOH is number of operating man-hours.l is tons of product.K is total number of process steps.P is capacity expressed in tons per day.

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The values of a are 23 (for a batch operation with maximum labor), 10 (for a well-instrumented continuous process operation), and 17 (for an operation with average labor requirements). Additional information on this method is available in Humphreys [32].

4.5.5 Cost Estimation Method V

This method is known as the turnover ratio method and is considered the most efficient approach to estimating plant costs. However, it is probably the least accurate. The turnover ratio is defined by [6,32]

TOR

ASI

=

(4.31)

whereTOR is turnover ratio.AS is gross annual sales. I is fixed capital investment.

The gross annual sales, AS, is expressed by

AS SP PR= ( )( ) (4.32)

whereSP is unit selling price.PR is yearly production rate.

Note that the value of the turnover ratio, TOR, usually varies from around 0.2 to 8.

Example 4.3Assume that a factory is to manufacture 50,000 units/year of a certain product. The selling price of a unit is $500. Calculate the fixed capital investment, if the turnover ratio is 4.

By substituting Equation (4.32) into Equation (4.31) and then substituting the given data values into the resulting equation, we get

4

500 50 000= ( )( , )I

(4.33)

By rearranging Equation (4.33), we obtain

I =

=

( )( , )

$ .

500 50 0004

6 25 million

Thus, the fixed capital investment for the factory is $6.25 million.

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Problems

1. Write an essay on life cycle cost models and cost estimation methods.

2. Discuss three types of life cycle cost models. 3. Write down life cycle cost equations for two general life cycle cost

models. 4. Write down life cycle cost equations for two specific life cycle cost

models. 5. Compare the general life cycle cost models with the specific life

cycle cost models. 6. Write down a life cycle cost equation for switching power supplies. 7. What is the “Boeing C-14 model”? 8. Discuss the following two types of cost estimation methods:

the Hand method•the Wessell method•

9. A solid-processing plant’s delivered equipment cost is $20 million. Calculate the total plant cost by using the Lang factor method.

10. An electric power generation company spent $1,500 million to construct a 600 MW nuclear power generating station. In order to meet the increasing demand for electricity, the company is plan-ning to construct a 1,500 MW nuclear power generating station. Calculate the cost of the new station, if the value of the cost-capac-ity factor is 0.7.

References

1. Dhillon, B. S. 1980. Life cycle cost: A survey. Microelectronics and Reliability: An International Journal 20:737–742.

2. Dhillon, B. S. 1983. Reliability engineering in system design and operation. New York: Van Nostrand Reinhold Company.

3. Gupta, Y. P. 1983. Life cycle cost models and associated uncertainties. In Electronic systems effectiveness and life cycle costing, ed. J. K. Skwirzyski, 535–549. Berlin: Springer–Verlag.

4. Dover, L. E., and Oswald, B. E. 1974. A summary and analysis of selected life cycle costing techniques and models. Master’s thesis, Air Force Institute of Technology, Wright-Patterson Air Force Base, Ohio.

5. Sherif, Y. S., and Kolarik, W. J. 1981. Life cycle costing concepts and practice. OMEGA 9:287–296.


7. Goldman, A. S., and Slattery, T. B. 1967. Maintainability: A major element of system effectiveness. New York: John Wiley & Sons.

8. Kolarik, W. J. 1977. Analysis theory and procedures for determining and pre-dicting availability, availability cost, and intangible effects for farm machinery systems. PhD dissertation, Oklahoma State University, Stillwater.

K10869_Book.indb 59 8/26/09 1:54:58 PM


9. Siewiorek, D. P., and Swarz, R. S. 1982. The theory and practice of reliable system design, digital press. Bedford, MA: Digital Equipment Corporation.

10. Reiche, H. 1980. Life cycle cost. In Reliability and maintainability of electronic sys-tems, ed. J. E. Arsenault and J. A. Roberts, 3–23. Potomac, MD: Computer Science Press, Potomac.

11. Locks, M. O. 1978. Maintainability and life cycle costing. Proceedings of the Annual Reliability and Maintainability Symposium 251–253.

12. Naval Weapons Engineering Support Activity, Naval Material Command. 1977. Life cycle cost guide for major weapon systems. Department of Defense, Washington, D.C.


14. Stordahl, N. C., and Short, J. L. 1968. The impact and structure of life cycle cost-ing. Proceedings of the Annual Symposium on Reliability 509–515.

15. Blanchard, B. S. 1978. Design and manage to life cycle cost. Portland, OR: M/A Press.

16. Department of the Army. 1976. Research and development cost guide for Army material systems. Pamphlet no. 11-12. Department of Defense, Washington, D.C.

17. Department of the Army. 1976. Investment cost guide for Army material sys-tems. Pamphlet no. 11-13. Department of Defense, Washington, D.C.

18. Department of the Army. 1976. Operating and support cost guide for Army material systems. Pamphlet no. 11-14. Department of the Army, Washington, D.C.

19. Monteith, D., and Shaw, B. 1979. Improved R, M, and LCC for switching power supplies. Proceedings of the Annual Reliability and Maintainability Symposium 262–265.

20. DeBurkarte, D. E. 1976. A standard life cycle cost model for inertial systems. Proceedings of the National Aerospace and Electronics Conference 687–695.

21. Ferens, D. V., and Harris, R. L. 1979. Avionics computer software operation and support cost estimation. Proceedings of the IEEE National Aerospace and Electronics Conference 296–300.

22. Boeing Aerospace Company. 1978. Advanced avionics systems for multimission applications. Report (vol. II)—Appendix G, Seattle, WA.

23. Desai, M. B. 1981. Preliminary cost estimating for process plants. Chemical Engineering July: 65–70.

24. Hackney, J. W. 1970. Estimating methods for process industry capital costs. In Modern cost-engineering techniques, ed. H. Popper, 43–58. New York: McGraw–Hill Book Company.

25. Ward, T. J. 1986. Cost-estimating methods. Modular instruction series G: Design of equipment (plant design and cost estimating), vol. 1, ed. J. Beckman, 12–21. New York: American Institute of Chemical Engineers.

26. Ostwald, P. F. 1974. Cost estimating for engineering and management. Englewood Cliffs, NJ: Prentice Hall, Inc.

27. Jelen, F. C., and Black, J. H., eds. 1983. Cost and optimization engineering. New York: McGraw–Hill Book Company.

28. Dieter, G. E. 1983. Engineering design. New York: McGraw–-Hill Book Company.

K10869_Book.indb 60 8/26/09 1:54:58 PM


29. Lang, H. J. 1947. Simplified approach to preliminary cost estimates. Chemical Engineering 54:130–133.

30. Hand, W. E. 1958. From flow sheet to cost estimate. Petroleum Refiner 37:331–334.

31. Wroth, W. F. 1960. Factors in cost estimation. Chemical Engineering 67:204–206. 32. Humphreys, K. K., ed. 1984. Project and cost engineers’ handbook, 51–74. New

York: Marcel Dekker, Inc.

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K10869_Book.indb 62 8/26/09 1:54:58 PM

63

5Reliability, Quality, Safety, and Manufacturing Costing

5.1 Introduction

Reliability, quality, safety, and manufacturing costs play an important role in the total cost of engineering products. Therefore, they must be considered with care. Reliability cost is an important factor in any reliability program associated with an engineering product. It is associated with activities such as reliability allocation, prediction, and testing [1].

Quality costs usually form a significant component of the selling price of an engineering product. They cross department lines by involving various company activities such as design, manufacturing, purchasing, and service. Safety costs are becoming an important element of the economy. For example, in 1995, the cost of workplace accidents in the United States was estimated to be around $75 billion [2]. Needless to say, safety costs are associated with areas such as lawsuits, insurance, analysis, and corrective measures.

The manufacturing cost may be described as the sum of fixed and variable costs chargeable to the manufacture of a given product or item. Usually, this cost (i.e., manufacturing cost) excludes the costs associated with corporate administration, selling, research and development, and transportation and distribution.

This chapter presents various important aspects of reliability, quality, safety, and manufacturing costing.

5.2 Reliability Cost Classifications

Reliability cost may be categorized under the following four classifications [3]:

Prevention cost• includes items such as hourly and overhead rates for design engineers, reliability engineers, material engineers, techni-cians, and test and evaluation personnel; hourly cost and overhead rates for reliability screens; cost of yearly reliability training per cap-ita; and cost of preventive maintenance programs.

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Appraisal cost• involves items such as cost for vendor audit, new ven-dor qualification, and new part qualification; hourly and overhead rates for reliability evaluation, reliability demonstration, reliability qualification, environmental testing, and life testing; cost of test result reports; and average cost per part of assembly testing, audit-ing, screening, inspection, and calibration.Internal failure cost• is composed of items such as cost of replaced parts or components; cost of spare part inventory; hourly and overhead rates for failure analysis, retesting, and troubleshooting and repair; and cost of production change administration.External failure cost• includes items such as cost of liability assurance, cost of warranty administration and reporting, cost of failure analy-sis, cost of spare part inventory, cost of service kit, cost of replaced parts, and cost to repair a failure.

5.3 Models for Estimating Costs of Reliability-Related Tasks

Over the years, many mathematical models have been developed to estimate man-hours required to perform reliability-related tasks and, in turn, the cost of performing such tasks. Some of these models are presented next [1,4,5].

5.3.1 Model I

This model is concerned with estimating the total number of man-hours required to perform reliability prediction. This number is expressed by [1,4,5]

TMHp = ( . )4 54 2 2α θ β (5.1)

whereTMHp is total number of man-hours required to perform reliability

prediction.a is the factor whose value depends on the type of report required: a = 1

means an internal report is required; a = 2 means a formal report is required.

q is the integer factor whose value varies from 1 to 3 depending on the level of detail: 1 = prediction exists, 2 = prediction is to be performed using similar system data, and 3 = full MIL-HDBK-217 [6] stress prediction is needed.

b is the integer factor whose values vary from 1 to 4 depending on the percentage of commercial hardware used in the system or item under consideration: 1 = 76–100%, 2 = 51–75%, 3 = 26–50%, and 4 = 0–25%.

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Reliability, Quality, Safety, and Manufacturing Costing 65

5.3.2 Model II

This model is concerned with estimating the total number of man-hours required to perform the reliability testing task. The number is defined by [1,4,5]

TMH HCFt = ( . )( )182 07 (5.2)

whereTMHt is the total number of man-hours required to perform the reliability

testing task.HCF is the integer factor whose value varies from 1 to 3 depending on the

degree of the hardware complexity: 1 = parts or components that are less than 15,000; 2 = parts or components that are 15,000–25,000; and 3 = parts or components that are greater than 25,000.

5.3.3 Model III

This model is concerned with estimating the total number of man-hours required for preparing the reliability and maintainability program plan. This number is defined by [1,4,5]

TMHpp = ( . )2 073 2γ (5.3)

whereTMHpp is total number of man-hours required to prepare the reliability

and maintainability program plan.g is number of MIL-STD-785/470 [6] tasks required. The recommended

minimum and maximum values of g are 4 and 22, respectively.

5.3.4 Model IV

This model is concerned with estimating the number of man-hours required for performing failure modes and effect analysis (FMEA). The number of man-hours required to perform this task is defined by [1,4,5]

TMH nf = ( . )17 79 (5.4)

whereTMHf is total number of man-hours needed to perform FMEA.n is total number of unique items requiring FMEA (e.g., number of circuit

cards for piece-part and circuit-level FMEA or the number of pieces of equipment for equipment-level FMEA). The recommended mini-mum and maximum values of n are 3 and 206, respectively.

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5.3.5 Model V

This model is concerned with estimating the total number of man-hours required to perform reliability allocation and modeling. The number of man-hours needed to carry out this task is defined by [1,4,5]

TMH CF Kam am= ( . )( )4 05 (5.5)

whereTMHam is total number of man-hours required to perform reliability allo-

cation and modeling.CFam is the allocation and modeling complexity. The recommended values

of the CFam are 1, 2, and 3 for a series system, simple redundancy, and very complex redundancy, respectively.

K is total number of items in the allocation process. The recommended minimum and maximum values of K are 7 and 445, respectively.

5.4 Quality Cost Classifications and Their Distribution in the Industrial Sector

Quality costs may be divided into four classifications, as shown in Figure 5.1 [7]: prevention cost, appraisal cost, internal failure cost, and external failure cost. Each of these classifications is described separately in the following sections.

Internalfailure

costPrevention

cost

Appraisalcost

Externalfailure

cost

Classifications

FIGURE 5.1Quality cost classifications.

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5.4.1 Prevention Cost

This cost is basically concerned with planning, implementing, and main-taining the quality system and is expressed by

C C C C Cp qe qt qp qd= + + + (5.6)

whereCp is prevention cost.Cqe is cost of quality engineering. This is concerned with the develop-

ment and implementation of the inspection plan, the overall qual-ity plan, etc.

Cqt is cost of quality training. This includes the cost of developing and maintaining quality-related training programs.

Cqp is cost of quality planning by functions excluding quality control.Cqd is cost associated with design and development of quality control and

measurement equipment.

5.4.2 Appraisal Cost

This cost is concerned with determining the degree of conformance to quality-related specifications. It has many elements, including the cost of conducting product quality audits, the cost of testing and inspection of incoming material, the cost of materials and services consumed in testing, the cost of inspection and testing of items being manufactured, and the cost of maintenance and calibration of equipment used for evaluating quality.

5.4.3 Internal Failure Cost

This cost occurs when manufactured items fail to meet specified quality requirements prior to their ownership transfer to customers. The subcatego-ries of the internal failure cost include repair cost, failure analysis cost, scrap cost, and re-inspection and retest cost.

5.4.4 External Failure Cost

This cost occurs when manufactured items fail to meet quality specifications after their delivery to customers. The external failure cost is expressed by

C C C C Cef r w a h= + + + (5.7)

whereCef is external failure cost.Cr is cost of repairing returned items.Cw is cost of warranties.Ca is cost of adjusting complaints.Ch is cost of replacement and handling of rejected (returned) items.

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Although the distribution of quality costs may vary from one industrial sector to another and from one organization to another, their distribution in the bank-ing and the electronic equipment manufacturing industries is as follows [8,9]:

Banking industry:• In this area, quality costs account for approximately 25% of a bank’s total operating costs. The estimates of their distribu-tion among prevention, appraisal, internal failure, and external fail-ure cost classifications are 2, 28, 41, and 29% (of the total quality cost), respectively.Electronic equipment manufacturing industry:• In this area, quality costs account for around 14% of the sales. The estimates of their distribu-tion among prevention, appraisal, internal failure, and external fail-ure cost classifications are approximately 45, 36, 13, and 6% (of the total quality cost), respectively.

5.5 Quality Cost Indexes and Quality Cost Reduction Approach

Many organizations use various types of quality cost indexes to monitor their performance. The values of such indexes are plotted on a periodic basis and their trends are monitored. Three of these indexes are presented next [10–13].

Index I is defined by

θ1

100100=

+( )( )C

Vq

o (5.8)

whereq1 is quality cost index.Cq is total quality cost.Vo is value of output.

The values of this index (q1) may be interpreted as follows [14]:

• q1 = 105 can readily be achieved in a real-life environment. • q1 = 110–130 occurs in companies where the quality costs are totally ignored. • q1 = 100 means that there is absolutely no defective output.

Index II is defined by

θ2

100=

( )( )C

Tq

s

(5.9)

where Ts is the total sales. Note that q2 is expressed as a percentage.

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Index III is defined by

θ3

100=

( )( )C

Cq

d

(5.10)

where Cd is the direct labor cost. Note that q3 is also expressed as a percent-age. Usually, this index is used to eliminate inflation effects.

Although quality costs can be reduced in many different ways, the six-step approach shown in Figure 5.2 is considered quite useful for this purpose [14]. Additional information on the approach is available in Williams [14].

5.6 Safety Cost and Its Related Facts and Figures

Nowadays, the cost of safety has become an important factor in the life cycle cost of many engineering systems. Each year, the cost of safety in general is increasing at a significant rate. Some of the safety cost–related facts and figures are as follows:

In 2000, work-related injuries cost the United States around $131 •billion [15].The cost of the accident in 1979 at the Three Mile Island nuclear •power plant was estimated to be approximately $4 billion [16].

Review past performance and existing conditions

Evaluate the environment

Develop objectives

Formulate and select a strategy

Implement the program

Report and review the plan and take necessary corrective measures

FIGURE 5.2A six-step approach for reducing quality costs.

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In 1993, a Virginia jury awarded $8 million to a worker for a back •injury caused by a piece of equipment that fell [16].In 1996, a Paris-bound Trans World Airlines jet crashed due to a fuel-•tank fire and killed all persons on board. A subsequent task force con-cluded that adding nonflammable gases (fuel-tank inverting) would decrease the risk of fuel-tank explosions quite significantly, but rec-ommended against such changes because of the cost of between $10 billion and $20 billion [17].In 1997, three workers sued a computer equipment manufacturer for •musculoskeletal disorders (MSDs) because they firmly believed that these disorders were due to keyboard entry activities [16]. The work-ers were awarded around $5.8 million.

5.7 Safety Cost Estimation Models

Over the years, many models to estimate safety cost have been developed. Some of these models are presented next.

5.7.1 Model I

This model is concerned with estimating the safety cost of a product over its life span and is expressed by [16,18]

LCSC C C C C Rp = + + + −1 2 3 4 (5.11)

whereLCSCp is product life cycle safety cost.C1 is cost of an accident prevention program.C2 is cost of insurance.C3 is recall cost.C4 is program cost.R is reimbursements.

5.7.2 Model II

This is another mathematical model that can also be used to estimate the safety cost of a product over its life span. This is expressed by [16,18]

LCSC SC SC SC SCp = + + +1 2 3 4 (5.12)

whereLCSCp is product life cycle safety cost.SC1 is safety cost during the product research and development phase.

This cost is associated with the safety-related studies performed dur-ing this phase.

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SC2 is safety cost during the product production and construction phase. This cost is associated with the safety-related measures taken during this phase.

SC3 is safety cost during the product operation and support phase. This cost is associated with safety-related activities performed during this phase.

SC4 is safety cost during the product retirement and disposal phase. This cost is associated with safety-related actions taken to dispose of the product.

5.7.3 Model III

This model is concerned with estimating the total hidden cost of an accident and is expressed by [18]

AHC C C C C C C C C C Cd m hsp uw iw e sp nw um ro= + + + + + + + + + (5.13)

whereAHC is total hidden cost of an accident.Cd is cost of damage to equipment or material.Cm is miscellaneous cost.Chsp is cost of time spent by clerical and higher supervisory personnel.Cuw is cost of wages paid to uninjured workers for the time lost.Ciw is cost of wages paid to injured workers for the time lost.Ce is extra cost of overtime work necessitated by the accident under

consideration.Csp is cost of wages paid to supervisory individuals for their time spent on

activities necessitated by the accident under consideration.Cnw is cost of the learning period required by new workers replacing

injured workers.Cum is uninsured medical cost borne by the organization or company.Cro is wage cost due to reduction in output of injured individuals after their

return to work.

5.7.4 Model IV

This model is concerned with estimating total safety cost, which is defined by [16,18]

C ILC PMC WIC IC IMC MIC LIC RRCst = + + + + + + + (5.14)

whereCst is total safety cost.ILC is cost of immediate losses due to accidents.PMC is cost of accident prevention measures.WIC is cost of welfare-related issues.IC is cost of insurance.IMC is cost associated with the immeasurable.

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MIC is cost of miscellaneous safety-related issues.LIC is cost of safety-related legal issues.RRC is rehabilitation and restoration cost.

5.8 Manufacturing Costs

Manufacturing costs form a significant proportion of the life cycle cost of engineering products, equipment, and systems. They may be broken down into five categories, as shown in Figure 5.3 [19,20]. For new processes, the ele-ments of the direct manufacturing cost include [19,20]:

maintenance and repair cost;•labor cost;•cost of utilities;•packaging and shipping cost;•raw materials cost;•royalties (if applicable);•direct overhead cost (usually plant supervision);•laboratory charges (process control and quality control work);•factory supplies (house supplies, wiping cloths, instrument charts, •etc.); anddevelopment cost (if applicable).•

Similarly, the elements of the indirect manufacturing cost are plant indirect overhead cost (e.g., plant office expense), property taxes, depreciation, and insurance [19,20].

Directmaterial cost

Directlabor cost

Indirectmaterial cost

Indirectlabor cost

Othermanufacturing expense

Manufacturingcost categories

FIGURE 5.3Manufacturing cost categories.

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5.9 Manufacturing Cost Estimation Models

Over the years, many mathematical models have been developed to esti-mate various types of manufacturing cost. Some of these models are presented next.

5.9.1 Model I

This model is concerned with estimating the direct cost of material used in manufacturing, which is expressed by [20,21]

C W P Pdm jj

s= +

−=

∑( )( ) 11

3

α (5.15)

whereCdm is direct material cost of a unit.W is weight of a unit, usually expressed in pounds.P is price of material expressed per linear foot, per pound, or per volume.aj is jth losses expressed in decimals for j = 1 (due to shrinkage), j = 2 (due

to scrap), and j = 3 (due to waste).Ps is unit price of expected material salvage expressed in dollars per unit.

Additional information on this model is available in Ostwald [21].

5.9.2 Model II

This model is concerned with estimating machining cost, which is expressed by [22–24]

MC

C r r rMT Tm

i=+

++

+1

601

1001100

1 2 3( ) ( )( ) (5.16)

whereMC is machining cost.Cm is machine cost expressed in dollars per hour.r1 is machine overhead rate expressed in percentage.r2 is operator labor rate expressed in dollars per hour.r3 is overhead rate of the operator expressed in percentage.MT is machining time.Ti is nonproduction or idle time.

Additional information on this model is available in references 22–24.

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5.9.3 Model III

This model is concerned with estimating the tool cost associated with a cut-ting tool brazed to the tool holder. This cost is defined by [22]

C

RSC TCt = +

+( )

( )β

β 1 (5.17)

whereCt is tooling cost associated with a cutting tool brazed to the tool holder.RSC is cost associated with resharpening expressed in dollars.b is number of resharpenings.TC is tool cost expressed in dollars.

In the case of a throwaway (insert) tool, the cost, Ct, is expressed by

C

THCn

TICnt = +

1 2

(5.18)

whereTHC is cost of the tool holder.n1 is total number of cutting edges in the life of the tool holder.TIC is tool insert cost expressed in dollars.n2 is number of cutting edges.

5.9.4 Model IV

This model is concerned with estimating the average unit cost for a single-point, rough-turning operation. This cost is expressed by [21]

C HC TC MC THCa = + + + (5.19)

whereCa is average unit cost for a single-point, rough-turning operation.HC is handling cost.TC is tool cost.MC is machining cost.THC is tool-changing cost.

Equations for estimating HC, TC, MC, and THC follow.The handling cost, HC, is expressed by

HC T OTCh= ( )( ) (5.20)

whereTh is total handling time per work piece expressed in minutes.OTC is total operating time cost expressed in dollars per minute.

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The tool cost, TC, is expressed by

TC

WPT CMTL

m e=( )( )

(5.21)

whereWPTm is machining time of work piece expressed in minutes per piece.Ce is tool cost expressed as dollars per cutting edge.MTL is mean tool life expressed in minutes.

The machining cost, MC, is expressed by

MC OTC WPTm= ( )( ) (5.22)

Finally, the tool-changing cost, THC, is expressed by

THC

OTC WPT TMTL

m c=( )( )( )

(5.23)

where Tc is the tool-changing time expressed as minutes per operation. Additional information on this model is available in Ostwald [21].

Problems

1. List and discuss reliability cost classifications. 2. Write an essay on reliability, quality, and safety costing. 3. What are the quality cost classifications? 4. Compare quality cost classifications with reliability cost

classifications. 5. Discuss an approach that can be used to reduce quality costs. 6. List at least five safety cost-related facts and figures. 7. List and discuss manufacturing cost categories. 8. Define at least two quality cost indexes. 9. Define a mathematical model that can be used to estimate the total

hidden cost of an accident. 10. Compare reliability cost with safety cost.

References

1. RADC reliability engineer’s toolkit. 1988. Published by the Systems Reliability and Engineering Division, Rome Air Development Center (RADC), Air Force Systems Command (AFSC), Griffiss Air Force Base, Rome, NY.

K10869_Book.indb 75 8/26/09 1:57:08 PM


2. Spellman, F. R., and Whiting, N. E. 1999. Safety engineering: Principles and prac-tice. Rockville, MD: Government Institutes.

3. Grant Ireson, W., and Coombs, C. F., eds. 1988. Handbook of reliability engineering and management. New York: McGraw–Hill Book Company.

4. National Technical Information Service (NTIS). 1987. R and M program cost drivers. Report no. RADC-TR-87-50 (ADA 182773), Springfield, VA.

5. Dhillon, B. S. 2005. Reliability, quality, and safety for engineers. Boca Raton, FL: CRC Press.

6. Department of Defense. 1998. Reliability program for systems and equipment. MIL-STD-785, Washington, D.C.

7. American Society for Quality Control. 1980. Guide for managing vendor quality costs. Milwaukee, WI.

8. Harrington, H. J. 1987. Poor-quality cost. New York: Marcel Dekker, Inc. 9. Breeze, J. D. 1980. Quality costs can be sold. Proceedings of the American Society for

Quality Control Conference 795–801. 10. Evans, J. R., and Lindsay, W. A. 1989. The management and control of quality. New

York: West Publishing Company. 11. Lester, R. H., Enrick, N. L., and Mottely, H. E. 1977. Quality control for profit.

New York: Industrial Press. 12. Carter, C. L. 1978. The control and assurance of quality, reliability, and safety.

Richardson, TX: C. L. Carter and Associates. 13. Sullivan, E., and Owens, D. A. 1983. Catching a glimpse of quality costs today.

Quality Progress 16 (12): 21–24. 14. Williams, R. J. 1982. Guide for reducing quality costs. Proceedings of the American

Society for Quality Control Annual Conference 360–366. 15. National Safety Council (NSC). 2001. Report on injuries in America in 2000.

Itasca, IL: Author. 16. Hammer, W., and Price, D. 2001. Occupational safety management and engineering.

Upper Saddle River, NJ: Prentice Hall, Inc. 17. Williams, W. E. 2001. Safety at all costs (www.worldnetdaily.com). Cave Junction,

OR, September 5: 1–3. 18. Dhillon, B. S. 2003. Engineering safety: Fundamentals, techniques, and applications.

River Edge, NJ: World Scientific Publishing. 19. Chemical Engineering (CE) cost file 92. 1964. Estimating manufacturing costs for

new processes. Chemical Engineering August 17: 160–162. 20. Dhillon, B. S. 1989. Life cycle costing: Techniques, models, and applications. New

York: Gordon and Breach Science Publishers. 21. Ostwald, P. F. 1974. Cost estimating for engineering and management. Englewood

Cliffs, NJ: Prentice Hall Inc. 22. Dieter, G. E. 1983. Engineering design: A materials and processing approach. New

York: McGraw–Hill Book Company. 23. Boothroyd, G. 1975. Fundamentals of metal machining and machine tools. New York:

McGraw–Hill Book Company. 24. Armarego, E. J. A., and Brown, R. H. 1969. The machining of metals. Englewood

Cliffs, NJ: Prentice Hall, Inc.

K10869_Book.indb 76 8/26/09 1:57:08 PM

77

6Maintenance, Maintainability, Usability, and Warranty Costing

6.1 Introduction

Each year billions of dollars are spent to produce various types of engi-neering products. Past experiences indicate that, in many cases, the cost of procuring an engineering product is less than the cost of ownership over its life span. According to Blanchard, Verma, and Peterson [1], the hidden costs related to equipment operation and support can account for as high as 75% of the equipment life cycle cost. Maintenance, maintainability, usabil-ity, and warranty costs play an important role in the life cycle cost of an engineering product. Therefore, careful consideration must be given to esti-mating such costs.

The maintenance cost may be described simply as the labor and materi-als expense required for maintaining engineering products in suitable use condition. In some systems—particularly military systems—the mainte-nance cost can be as high as 70% of life cycle costs [2]. Maintainability is an important factor in the total cost of equipment because increase in maintainability can result in reduction in equipment operation and support costs. Thus, maintainability costs are basically concerned with equipment design.

Usability costs are concerned with a wide range of activities employed in developing effectively usable engineering products. Some examples of these activities are establishing a definition for end user requirements, developing specifications for usability objectives, performing task analysis, and conducting usability testing. Warranty costs occur when engineering equipment manufacturers provide buyers with written statements guaran-teeing the integrity of their equipment. The responsibilities of the manufac-turers are outlined by these statements in situations when their equipment happens to be defective.

This chapter presents various important aspects of maintenance, main-tainability, usability, and warranty costing.

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6.2 Reasons for Maintenance Costing, Factors Influencing Maintenance Cost, and Types of Maintenance Costs

There are many reasons for maintenance costing. Some of the important ones include [3]:

to prepare budgets;•to make equipment replacement decisions;•to control costs;•to compare maintenance costs’ effectiveness with industry averages;•to identify maintenance cost drivers;•to improve productivity;•to compare competing maintenance methods;•to provide appropriate inputs in the design of new equipment or •items; andto perform equipment or item life cycle cost studies.•

Some of the important factors influencing maintenance costs are shown in Figure 6.1 [3,4].

Influencingfactors

Operational environment

Regulatorycontrols

Equipment specification

Company policy

Type of service

Maintenancepersonnel

skills

Operatorexpertise and

experience

Asset condition(i.e., age, condition,

and type)

FIGURE 6.1Factors influencing maintenance costs.

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Maintenance, Maintainability, Usability, and Warranty Costing 79

There are basically two main categories of maintenance costs: preventive maintenance cost and corrective maintenance cost. The former is concerned, directly or indirectly, with actions performed on a planned, periodic, and specific schedule for keeping a piece of equipment or item in stated working condition through the process of rechecking and reconditioning. More spe-cifically, these actions are precautionary measures undertaken to forestall or decrease the probability of failures or an unacceptable level of degradation in subsequent service, rather than rectifying failures after their occurrence.

The corrective maintenance cost is directly or indirectly concerned with the unscheduled maintenance and repair to return equipment or items to a specified condition. These actions are carried out because involved mainte-nance personnel or users perceive deficiencies or failures.

6.3 Equipment Maintenance Cost

The maintenance cost of the entire ownership cycle of equipment is expressed by [4,5]

EMC CMC PMC

jjp c p

m

= +− +

−[ ( ) ( )]

( )λ λ 1 1 (6.1)

whereEMCp is present value of the maintenance cost of the entire ownership

cycle of equipment.l c is constant corrective maintenance rate of equipment per year.CMC is expected cost of a corrective maintenance action.lp is constant preventive maintenance rate of equipment per year.PMC is expected cost of a preventive maintenance action.m is equipment expected life expressed in years.j is annual interest rate.

Example 6.1Assume that annual preventive and corrective maintenance rates of an engineering system are 5 and 2, respectively. Each preventive and corrective action costs $200 and $1,000, respectively. Calculate the present value of the system maintenance cost, if the expected system life and annual interest rate are 10 years and 5%, respectively.


EMCp = +− +

−[( ) ( , ) ( ) ( )]

( . ).

2 1 000 5 2001 1 0 05

0 05

10

= $ , .23 165 20

Thus, the present value of the system maintenance cost is $23,165.20.

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6.3.1 Maintenance Equipment Cost

This is expressed by [6]

MEC C Crd a= +α (6.2)

whereMEC is maintenance equipment cost.Crd is research and development cost associated with the maintenance

equipment.a is number of pieces of maintenance equipment.Ca is maintenance equipment unit acquisition cost.

6.4 Preventive and Corrective Maintenance Labor Cost Estimation

The preventive maintenance labor cost is expressed by [7]

PMLC LRf APMT

f

j jj

K

jj

K=

=

=

∑∑

( ) 1

1

(6.3)

wherePMLC is equipment preventive maintenance labor cost.LR is hourly labor rate.K is number of data points.fj is frequency of jth preventive maintenance action expressed in actions

per operating hour, after adjustment for equipment duty cycle, for j = 1, 2, 3,…, K.

APMTj is average time, in hours, required to carry out jth preventive main-tenance action for j = 1, 2, 3,…, K.

Similarly, the corrective maintenance labor cost is given by [7]

CMLC

T LC MTTRMTBF

so=( )( ) (6.4)

whereCMLC is equipment annual corrective maintenance labor cost.Tso is equipment annual scheduled operating hours.LC is hourly corrective maintenance labor cost.MTTR is equipment mean time to repair.MTBF is equipment mean time between failures.

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Example 6.2Assume that a system is scheduled to operate for 2,500 hours annually and its mean time between failures and mean time to repair are 700 hours and 3 hours, respectively. Calculate the system annual corrective maintenance labor cost, if the hourly corrective maintenance labor cost is $30.

By inserting the given data values into Equation (6.4), we get

CMLC =

=

( ) ( ) ( )

$ .

2500 30 3700

321 43

Thus, the system annual corrective maintenance labor cost is $321.43.

6.5 Repair Manpower, Maintenance Material, and Spare and Repair Parts Costs

According to a U.S. military document [6], repair cost with respect to man-power can be estimated by using the following equation:

RMC F RCrs um= −θ ( )1 (6.5)where

RMC is repair manpower cost.q is total number of repairable units failing over system or equipment life.Frs is repairable shrinkage factor due to damage, loss, etc. Its values are

tabulated in reference 6 and vary from 0 to 0.1375.RCum is unit repair cost with respect to manpower.

The total number of repairable units failing over system or equipment life is expressed by

θ λ α= ( )( )( )T SL (6.6)where

l is item constant failure rate.T is annual operating hours.a is number of repairable items.SL is system or equipment life (in reference 6, taken to be 10 years).

The unit repair cost with respect to manpower is given by

RC x F yum mu= ( )( )( ) (6.7)where

x is mean number of man-hours per repair action.Fmu is manpower use factor. Its values are tabulated in reference 6 and they

vary from 1.04 to 3.y is hourly manpower cost including overhead.

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The maintenance material cost is an important element of the total mainte-nance cost. For example, according to Neibel [8], in the United States indus-trial sector, the cost of maintenance materials typically accounts for 40–50% of the total maintenance cost. Because the cost of excessive inventory and obsolete parts is an important factor in most maintenance stockrooms and storerooms, well-planned and efficiently operated stockrooms and store-rooms can help to reduce the cost of materials.

The total cost of stock or stores at the time of repair can be calculated by using the following equation [8]:

SC ITC C W ITC ITC t ITCt i it= + + − + +

=

( ) ( . )( )( ) ( . )( )0 01 0 1

IITC Ct ITC ITC

i+ + +

( )( ) ( )( )10100

(6.8)

whereSCt is total cost of stock or stores at the time of repair.Ci is inventory cost per item.ITC is present worth of the inventory item cost including procurement and

delivery costs.Wit is inventory item worth after K periods.t is time, expressed in months, during which the stock item is in inventory.

Note that Equation (6.8) allows an inflation rate of 1% per month of procure-ment cost, while the item under consideration is in inventory, and 10% for the item’s total shelf life to take into consideration factors such as spoilage, obsolescence, theft, and deterioration.

Equations to calculate Wit, Ci, and ITC, respectively, are presented next.

W ITC jitK= +( )( )1 (6.9)

C S FSC n yi b= ( )( )/( )( ) (6.10)

ITC UL SL PC WT MSP= + + −( )( )( )1 (6.11)

wherej is interest rate for a given period.K is total number of interest periods.Sb is size of a bin expressed in square feet.FSC is yearly floor space cost per square foot.n is mean number of items stored in a bin.y is reciprocal of total years that an item usually spends in inventory.UL is amount of losses generated by unused stock returned to inventory

considered too small in terms of quantity for use in the future.

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SL is total amount of losses due to scrap, chips, skeletons, and so on.PC is procurement cost or price (i.e., the delivered price) of material per

unit.WT is weight or other unit of quantity of material used.MSP is unit price of material salvaged.

The spare and repair parts cost is defined by

SRC ISRC CC DSRC SSRC OSRC= + + + + (6.12)

whereSRC is spare and repair parts cost.ISRC is intermediate spare and repair parts cost.CC is total cost of consumables.DSRC is depot spare and repair parts cost.SSRC is supplier spare and repair parts cost.OSRC is organizational spare and repair parts cost.

6.6 Maintenance Cost Estimation Models

Over the years, many mathematical models have been developed to estimate various types of maintenance-related costs. Four of these models are pre-sented next.

6.6.1 Model I

This model is concerned with estimating equipment initial logistic support cost, which is expressed by [9]

EILSC ISRC ITHC IMC TDPC LPMC ITTEC PC OTSEPC= + + + + + + + (6.13)

whereEILSC is equipment initial logistic support cost.ISRC is cost of initial spare and repair parts.ITHC is initial transportation and handling cost.IMC is cost of initial inventory management.TDPC is technical data preparation cost.LPMC is cost of logistic program management.ITTEC is initial training and training equipment cost.PC is provisioning cost, including preparation of procurement data for

essential spares, test, and support equipment.OTSEPC is procurement cost of operational test and support equipment.

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6.6.2 Model II

This model is concerned with estimating software maintenance cost. This cost is expressed by [10]

SMC

m LC= 3( )( )θ (6.14)

whereSMC is software maintenance cost.m is total number of instructions to be changed per month.LC is labor cost per man-month.q is difficulty constant. Its values for hard programs, easy programs, and

programs of medium difficulty are 100, 500, and 250, respectively.

6.6.3 Model III

This model is concerned with estimating Doppler radar maintenance cost. This is expressed by [11]

DRMC

C ya=( )1000

(6.15)

whereDRMC is Doppler radar maintenance cost.Ca is Doppler radar annual maintenance cost.y is number of years in service.

The natural logarithm of Ca is given by

l nC l nCa fu= +θ θ1 2 (6.16)

whereq1 = –1.269q2 = 0.696Cfu is the first unit cost of the Doppler radar expressed in 1974 dollars (× 103).

6.6.4 Model IV

This model is concerned with estimating the fire control radar maintenance cost, which is expressed by [11]

FCRMC MC h hph= ( ) /1 2 1000

(6.17)

whereFCRMC is fire control radar maintenance cost.MCph is maintenance cost per flying hour per unit expressed in 1974

dollars (× 103).h1 is total number of annual flying hours.h2 istotal number of years in service.

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The natural logarithm of MCph is expressed by

ln MC ln Pph pw= +α α1 2 (6.18)

wherea1 = –2.086a2 = 0.611Ppw is peak power expressed in kilowatts.

6.7 Maintenance Cost Data Collection

As various types of cost data are needed in maintenance costing, manage-ment decides the types of cost data the maintenance department should col-lect by considering their potential applications. Four types of maintenance cost-related data are collected [12]:

Labor costs• are usually obtained by using items such as timesheets, job tickets, and maintenance work orders.Spare parts and supplies costs• are usually obtained from maintenance work orders.Overhead costs• are usually obtained from the company accounting department.Equipment costs• are usually obtained from either purchase orders or suppliers’ invoices.

6.8 Maintainability Investment Cost Elements

The main elements of maintainability investment cost are as follows [6]:

repair parts;•system test and evaluation;•new operational facilities;•system engineering management;•data;•training;•prime equipment; and•support equipment.•

Additional information on these elements is available in reference 6.

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6.9 Manufacturer Warranty and Reliability Improvement Warranty Costs

The cost of warranty to an equipment manufacturer can be quite significant. It can be estimated by using the following equation [13]:

MWC C n Cmu fw= +( )( )( )λ (6.19)

whereMWC is manufacturer or contractor warranty cost.Cmu is mean cost for the manufacturer or contractor to repair a unit sent

back for warranty service.n is operating hours of equipment under warranty during the warranty

period.l is average constant failure rate per hour of equipment under warranty

during the warranty period. Cfw is manufacturer or contractor warranty fixed cost.

The manufacturer warranty and reliability improvement warranty cost is expressed by [14]

MWRC FC C C P Cm ia d x= + + + + (6.20)

whereMWRC is manufacturer warranty and reliability improvement warranty

cost.FCm is fixed cost of the manufacturer associated with the warranty.Cia is cost associated with reliability improvement actions for attaining the

achieved mean time between failures (i.e., reliability improvement warranty period average).

Cd is cost of damages associated with not meeting the specified turnaround time.

P is profit.Cx, is cost.

The cost, Cx, is expressed by

C C n T UR MTBFx mr w a= ( )( )( )( )/ (6.21)

whereCmr is manufacturer’s cost per unit repair.n is number of systems or items to be delivered.Tw is length of the warranty period.UR is usage rate expressed in operating time per calendar time.MTBFa is achieved mean time between failures (i.e., reliability improve-

ment warranty period average).

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6.10 Usability Costing and Related Facts and Figures

A wide range of activities is generally employed in effectively developing usable engineering products. The cost of these activities depends on factors such as the scope of the product under consideration, functional range, the number of scenarios to be studied, the number of users to be studied, and the skill and experience of the usability specialists [15]. Some of the usability costing-related facts and figures include:

An American Airlines study reported that catching a usability-related •problem early in the design process can decrease the cost of correct-ing it by 60–90% [16].A study revealed that the total training time for new users of a stan-•dard personal computer was approximately 21 hours as opposed to around 11 hours for users of a user-friendly computer [17].A study reported that the annual cost of lost productivity to •American businesses is around $100 billion because office workers “futz” with their machines an average of 5.1 hours per week [18].A study revealed that approximately 80• % of software maintenance cost is due to unmet or unforeseen user requirements [19].A study reported that an Australian insurance company spent •approximately $100,000 (Australian) on a usability-related project concerned with redesigning its application forms to reduce customer errors and saved $536,023 (Australian) annually [20].

Additional information on usability-related facts and figures is available in Dhillon [21].

6.11 Principal Costs of Ignoring Product Usability and Product Usability Cost Estimation

The principal costs of ignoring engineering product usability are as follows [22]:

User error cost• is concerned with the users of engineering prod-ucts making errors. In turn, these errors result in reduction in their productivity.Poor productivity cost• consists of the additional time spent by engi-neering product users with products that are difficult to use.Training cost• deals with the training of users when the product is first introduced. It increases significantly when products are difficult to use.

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Customer support cost• involves a customer hotline telephone service, usually provided by product manufacturers for people having dif-ficulties using the product. Past experiences indicate that products that are difficult to use generate greater customer or user requests for help. In turn, more people are required to handle users or cus-tomers, thus resulting in greater customer support cost.Poor sale cost• involves dissatisfied customers or users not purchas-ing the product in the future, even when they are made aware of improvements in product usability. Past experiences indicate that a dissatisfied customer or user influences roughly 10 others to avoid buying the product in question [22].Tarnished corporate image cost• is concerned with users or customers buying not only the current or improved usability version of the product in question, but also other products manufactured by the same firm.

The product usability cost can be estimated by using the following equa-tion when usability cost data are available for similar products of different capacities [23]:

DPC SPC

CPCPu u

d

o

=

θ

(6.22)

whereDPCu is desired product usability engineering cost.SPCu is known usability engineering cost of a similar item, product, or

piece of equipment of known capacity CPo.CPd is desired product capacity.q is cost-capacity factor. The value of this factor varies for different prod-

ucts or items. In circumstances when no data for q are available, it is considered quite reasonable to assume its value to be 0.6.

Example 6.3Assume that the usability engineering cost of an 80 GB computer system is $300. Calculate the cost of usability engineering of a similar 100 GB computer system if the value of the cost-capacity factor is 0.8.


DPCu =

=

( )

$ .

.

30010080

358 63

0 8

Thus, the cost of usability engineering of the similar 100 GB computer system is $358.63.

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Problems

1. What are the principal reasons for maintenance costing? 2. Discuss at least seven factors that influence maintenance cost. 3. Assume that annual preventive and corrective rates of an engi-

neering system are 7 and 3, respectively. Each preventive and cor-rective action costs $400 and $1,200, respectively. Calculate the present value of the system maintenance cost, if the expected sys-tem life and annual interest rate are 12 years and 3%, respectively.

4. Discuss the collection of three types of maintenance cost-related data.

5. What are the principal elements of maintainability investment cost?

6. Discuss equipment warranty cost. 7. What is the equipment usability cost? 8. List at least five usability costing-related facts and figures. 9. Discuss the costs of ignoring engineering product usability. 10. Assume that an engineering system is scheduled to operate for

3,000 hours annually and its mean time between failures and mean time to repair are 800 hours and 4 hours, respectively. Calculate the system annual corrective maintenance labor cost, if the hourly corrective maintenance labor cost is $40.

References

1. Blanchard, B. S., Verma, D., and Peterson, E. L. 1995. Maintainability: A key to effective serviceability and maintenance management. New York: John Wiley & Sons.


3. Levitt, J. 1997. The handbook of maintenance management. New York: Industrial Press.

4. Dhillon, B. S. 2002. Engineering maintenance: A modern approach. Boca Raton, FL: CRC Press.

5. Dhillon, B. S. 1996. Engineering design: A modern approach. Chicago, IL: Richard D. Irwin.

6. Department of the Army. 1976. Engineering design handbook: Maintainability engineering theory and practice, AMCP 706-133, Department of Defense, Washington, D.C.

7. Department of the Army. 1975. Engineering design handbook: Maintenance engi-neering techniques. AMCP 706-132. Department of Defense, Washington, D.C.

8. Neibel, W. B. 1994. Engineering maintenance management. New York: Marcel Dekker.

K10869_Book.indb 89 8/26/09 1:59:28 PM


9. Dhillon, B. S. 1999. Engineering maintainability. Houston, TX: Gulf Publishing Company.

10. Sheldon, M. R. 1979. Life cycle costing: A better method of government procurement. Boulder, CO: Westview Press.

11. Cost analysis of avionics equipment, vol. 1. 1974. Prepared by U.S. Air Force Systems Command, Wright-Patterson Air Force Base, Ohio. The NTIS report no. AD741132. Available from the National Technical Information Service (NTIS), Springfield, VA.

12. Jordan, J. K. 1990. Maintenance management. Denver, CO: American Water Works Association.

13. Balaban, H. S., and Meth, M. A. 1978. Contractor risk associated with reliability improvement warranty. Proceedings of the Annual Reliability and Maintainability Symposium 123–129.

14. Gates, R. K., Bicknell, R. S., and Bortz, J. E. 1977. Quantitative models used in the RIW decision process. Proceedings of the Annual Reliability and Maintainability Symposium 229–236.

15. Rosson, M. B., and Carroll, J. M. 2002. Usability engineering: Scenario-based development of human-computer interaction. San Francisco: Morgan Kaufmann Publishers.

16. Laplante, A. 1992. Put to the test. Computerworld 27 (July 27): 75–77. 17. Nielson, J. 1993. Usability engineering. Boston: Academic Press, Inc. 18. Westlake Consulting Company. 1997. SBT Accounting Systems. Houston, TX. 19. Pressman, R. S. 1992. Software engineering: A practitioner’s approach. New York:

McGraw–Hill Book Company. 20. Fisher, P., and Sless, D. 1990. Information design methods and productivity in

the insurance industry. Information Design Journal 6 (2): 103–129. 21. Dhillon, B. S. 2004. Engineering usability: Fundamentals, applications, human fac-

tors, and human error. Stevenson Ranch, CA: American Scientific Publishers. 22. Keinonen, T., Mattelmaki, T., Soosalu, M., and Sade, S. 1997. Usability design

methods. Technical report, Department of Product and Strategic Design, University of Art and Design, Helsinki, Finland.

23. Dieter, G. E. 1983. Engineering design: A materials and processing approach, 324–366. New York: McGraw–Hill Book Company.

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91

7Computer System Life Cycle Costing

7.1 Introduction

Today computers play an important role in our daily lives. Over the years, their applications have increased quite dramatically, ranging from personal use to controling nuclear reactors and space systems. The computer industry has become an important component of the global economy; a vast sum of money is spent to produce, operate, and maintain computers each year. For example, in fiscal year 1980, the U.S. government spent over $57 billion on computer systems [1].

Computer systems are made up of both hardware and software compo-nents and the percentage of overall computer system cost spent on hardware has changed quite remarkably over the years. For example, in 1955, the hard-ware component accounted for 80% of total computer system cost; however, in 1985, the hardware component cost decreased to just 10% [2]. This means that, nowadays, software cost is a very important element of total computer system cost—more specifically, the computer system life cycle cost.

Over the years, many models and procedures have been developed to esti-mate directly or indirectly computer system life cycle cost. This chapter pres-ents various important aspects of computer system life cycle costing.

7.2 Computer System Life Cycle Cost Models

A number of mathematical models are used to estimate life cycle cost of a computer system. Two such models are presented next [3,4].

Model I divides the life cycle cost of a computer system into two main com-ponents: procurement cost and ownership cost. Thus, the life cycle cost of a computer system is expressed by

LCC C CCS p1 0= +

(7.1)

whereLCCCS1 is computer system life cycle cost.Cp is computer system procurement cost.C0 is computer system ownership cost.

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The procurement cost includes the cost of items such as system hardware, software license fees, installation, training, and documentation. Similarly, the ownership cost includes the cost of items such as preventive mainte-nance, computer downtime, supplies, and corrective maintenance.

Model II is a more detailed model to estimate the life cycle cost of a com-puter system. However, the model assumes that the cost of corrective main-tenance is the only ownership cost of the computer system. Thus, the life cycle cost of the computer system is defined by

LCC C C iCS p mj jj

j

n

2 21

1= + +=

∑ ( ) /( )α (7.2)

whereLCCCS2 is life cycle cost of the computer system.Cp2 is procurement cost of the computer system.n is computer system expected life expressed in years.Cmj is corrective maintenance cost of a single maintenance activity during

year j.aj is expected number of times that the computer system will fail during

year j.i is discount rate.

For the same number of computer system failures occurring in each year, Equation (7.2) simplifies to

LCC C C iCS p mjj

j

n

2 21

1= + +=

∑α /( ) (7.3)

where a is expected number of computer system failures per year.

Example 7.1Assume that the procurement cost and expected useful life of a computer system are $4,000 and 5 years, respectively. The computer system’s expected number of failures per million hours is 100 and its only ownership cost is the cost of correc-tive maintenance. Calculate the life cycle cost of the computer system, if the cost of each corrective maintenance call is $200 and the yearly discount or interest rate is 4%.

The expected number of failures of the computer system per year is given by

n failures year= =

( ) ( , ), ,

. /100 8 7601 000 000

0 876

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Computer System Life Cycle Costing 93

Using the preceding calculated value and the given data in Equation (7.3), we get

LCCCSj

j2

1

5

4 000 0 876 200 1 0 04

4 779

= + +

=

−

=∑, ( . ) ( ) ( . )

$ , ..96

Thus, the life cycle cost of the computer system is $4,779.96.

7.3 Computer System Maintenance Cost

The computer system maintenance cost is an important component of com-puter system life cycle cost. This section presents two mathematical models to estimate, directly or indirectly, computer system maintenance cost.

Model I is concerned with estimating the maintenance cost of computer system hardware, which is expressed by [3,4]

Csm = Cpm + Ccm + Ci (7.4)

whereCsm is monthly maintenance cost of the computer system hardware.Cpm is preventive maintenance cost of the computer system hardware.Ccm is corrective maintenance cost of the computer system hardware.Ci is inventory cost.

The preventive maintenance cost of the computer system hardware is expressed by

C OH SPMT TT SPMpm e e i= +( ) [ ]/θ (7.5)

whereOH is equipment operating hours per month.q is hourly rate of the customer engineer. This also includes the spare parts

usage rate.SPMTe is customer engineer’s scheduled preventive maintenance time.TTe is customer engineer’s travel time to perform preventive maintenance.SPMi is scheduled preventive maintenance interval.

The customer engineer’s hourly rate, q, is expressed by

θ

α= + +PR OR

Cp

( )1 (7.6)

K10869_Book.indb 93 8/26/09 2:00:10 PM


wherePR is hourly pay rate of the customer engineer.OR is overhead rate.Cp is cost of parts per hour.a is fraction of time that the customer engineer spends on the maintenance

activity. Note that the customer engineer spends the remaining frac-tion of time on items such as paperwork, training, and waiting.

The corrective maintenance cost of the computer system hardware is defined by

C OH TT MTTR MTBFcm ecm= +θ ( ) [ ]/ (7.7)

whereTTecm is customer engineer’s travel time for performing corrective

maintenance.MTTR is mean time to repair.MTBF is mean time between failures.

The inventory cost, Ci, is expressed by

C V Ri ip i= ( )( ) (7.8)

whereVip is value of the maintenance spare parts inventory.Ri is monthly inventory cost rate, which includes items such as handling

cost, interest charges for spares, and depreciation.

Model II is concerned with estimating the annual labor cost of servicing a computer system. This cost depends on many factors, including average cost of labor, mean time to preventive maintenance, preventive maintenance time interval, mean time between failures, and mean time to repair.

The annual labor cost is defined by

C LCT TT

TTT MTTR

MTBFa hapm pm

bpm

r=+

++

( )( )( ) ( )

8760

(7.9)

whereLCh is labor cost per hour.Tapm is average time taken to perform preventive maintenance.TTpm is travel time associated with a preventive maintenance call.Tbpm is mean time between preventive maintenance services.TTr is travel time associated with a repair or corrective maintenance call.MTTR is mean time to repair. MTBF is mean time between failures.

K10869_Book.indb 94 8/26/09 2:00:29 PM


Example 7.2Assume the following data values concerning servicing a computer system:

LCh = $50Tapm = 4 hoursTTpm = 0.5 hourTbpm = 2,500 hoursTTr = 1 hourMTTR = 2 hoursMTBF = 3,000 hours

Calculate the annual labor cost for servicing the computer system by using Equation (7.9).


Ca = + + +

=

( ) ( )( . )

,( ),

$ ,

50 87604 0 52 500

1 23 000

1 2226 4.

Thus, the annual labor cost for servicing the computer system is $1,226.40.

7.4 Software Costing and Related Difficulties

Over the years, software cost has increased to a very high level from a rather low percentage of the total computer system cost. For example, according to a U.S. Air Force study conducted in 1972, cost of software in 1955 accounted for less than 20% of total computer system cost; how-ever, its projection for 1985 was around 80% of the total amount [3]. Furthermore, in July 1976, Newsweek magazine reported that the ratios of computer system hardware cost to software cost were 1:4 and 4:1 in 1976 and the 1950s, respectively.

Needless to say, today software cost has become a very important element of the computer system life cycle cost. Over time, many methods and proce-dures have been developed to estimate software cost.

Some of the difficulties faced in estimating software cost include [5]:

poor understanding of the effects of management and technical-•related constraints;poor understanding of the software development and mainte-•nance processes;unavailability of adequate historic data to make appropriate •checks;

K10869_Book.indb 95 8/26/09 2:00:35 PM


unavailability of adequate historic data for calibration applications •(A calibration may be described as a process through which a model is fitted to a given cost estimating condition.); andproject-to-project comparison inhibition because of firm belief in a •project’s uniqueness.

7.5 Software Life Cycle Cost Influencing Factors and Model

Many factors influence software life cycle cost. They may be grouped under five distinct attributes [6]:

Group I: computer attributes.• Some examples of these attributes are turnaround time, speed, and storage constraints.Group II: project attributes.• Some examples of these attributes are schedule constraints, use of software tools, and modern program-ming practices [7].Group III: size attributes.• Some examples of these attributes are num-bers of inputs, outputs, data elements, and instructions.Group IV: product attributes.• Some examples of these attributes are required software reliability, the choice of programming language, and software product complexity.Group V: personnel attributes.• These attributes affect software cost much more than any other groups of attributes. Some examples of personnel attributes are teamwork; experience with respect to items such as programming language, applications, and virtual machines; and personnel and team capabilities.

The life cycle cost of software is composed of seven distinct elements, as shown in Figure 7.1. This is expressed mathematically by [8]

LCC SDC SAC CC SOSC SIC STIC SDOCse = + + + + + + (7.10)

whereLCCse is software life cycle cost.SDC is software design cost.SAC is software analysis cost.CC is software code and checkout cost.SOSC is software operating and support cost.SIC is software installation cost.STIC is software test and integration cost.SDOC is software documentation cost.

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Table 7.1 presents the main elements of costs of software design, analysis, operating and support, code and checkout, test and integration, installation, and documentation [8].

7.6 Software Cost Estimation Methods and Models

Over the years, many methods and models have been developed to estimate software cost. Some of these methods and models are presented next.

7.6.1 Software Cost Estimation Methods

Many methods have been used to estimate software costs, including [3,9]:

algorithmic models;•top-down estimating;•bottom-up estimating;•analogy; and•expert opinion.•

The algorithmic models are described later in detail, and additional informa-tion on the remaining four methods is available in Dhillon [3] and Boehm [9].

The algorithmic models may be described as the models that provide at least one mathematical algorithm to generate a computer software cost estimate as a

Code and checkout

cost

Operatingand support

cost

Documentationcost

Design cost

Analysiscost

Test and integration

cost

Installation cost

Software life cycle cost elements

FIGURE 7.1Software life cycle cost elements.

K10869_Book.indb 97 8/26/09 2:00:39 PM


function of several variables. These variables are considered very important cost drivers. The five common types of algorithmic models are presented next [9].

7.6.1.1 Tabular Models

Generally, these models are quite straightforward to comprehend and imple-ment. They are composed of tables relating cost driver variables’ values to portions of the software development effort or to multipliers employed to adjust the effort estimate. Three examples are the Black et al. [10], Aron [11], and Wolverton [12] models.

7.6.1.2 Composite Models

These models incorporate an amalgamation of four types of functions (i.e., linear, tabular, analytic, and multiplicative) for determining software effort

TABLE 7.1

Subelements of Software Life Cycle Cost Elements

No.Software Life Cycle

Cost Element Subelements

1 Design cost Cost of flow chartsData structure costCost of test proceduresCost of input and output parameters

2 Analysis cost Cost of system requirementsCost of program requirementsCost of design requirements and specificationsCost of interface requirements

3 Operating and support cost Cost of modificationsCost of test revisionsCost of documentation revisionsCost of environments

4 Code and checkout cost Cost of desk checksCost of coded instructionsCost of compiling programs

5 Test and integration cost Cost of program testCost of system integration

6 Installation cost Validation costVerification costCertification cost

7 Documentation cost Cost of listingsCost of user manualCost of maintenance manual

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as a function of cost driver variables. Past experiences indicate that compos-ite models are relatively more difficult to learn and use, in addition to requir-ing more data and effort.

7.6.1.3 Analytic Models

These models take the following form [3]:

ET f x x xn= ( , , ...., )1 2 (7.11)

whereET is effort.f is a function (it is to be noted that this function is neither linear nor

multiplicative).xj is cost driver variable j for j = 1, 2, 3,…, n. n is total number of cost driver variables.

Two good examples of the analytic models are the Putnam [13] and Halstead [14] models.

7.6.1.4 Linear Models


ET N N xo j jj

K

= +=

∑1

(7.12)

whereET is effort.K is total number of cost driver variables.Nj is coefficient chosen to best fit the observed data points for j = 0, 1, 2,

3,…, K.xj is cost driver variable j for j = 1, 2, 3,…, K.

An important reference to the earliest use of the linear model is the System Development Corporation software cost-estimation study performed in the mid-1960s [15]. Finally, it was concluded that there are many nonlinear inter-actions in the software development process for linear models to perform very effectively.

7.6.1.5 Multiplicative Models


ET M Mo j

x

j

nj=

=∏

1

(7.13)

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whereET is effort.n is total number of cost driver variables.Mj is coefficient chosen to best fit the observed data for j = 0, 1, 2, 3,…, n. xj is cost driver variable j for j = 1, 2, 3,…, n.

Past experiences indicate that multiplicative models work fairly well for reasonable, independently selected variables. Additional information on multiplicative models is available in Walston [15] and Herd et al. [16].

7.6.2 Software Cost Estimation Models

As mentioned earlier, many types of models can be used to estimate soft-ware cost (see references 3, 5, 9, and 17). Some of these models used directly or indirectly to estimate software cost are presented next.

Model I is concerned with estimating the software development cost, which is expressed by [16,18]

SDC SPC SSC= + (7.14)

whereSDC is software development cost.SPC is software primary development cost.SSC is software secondary development cost.

The software primary development cost is defined by

SPC ALR MR= ( )( ) (7.15)

whereALR is average labor rate of the software development manpower expressed

in dollars per man-month. It includes items such as administration cost, general cost, and overhead cost.

MR is manpower required for software development expressed in man-months. This includes activities such as analysis, design, code, test, debug, and checkout.

Similarly, the software secondary development cost is expressed by

SSC C

SPC

ALR MR

ii

m

=

=

=

=∑

1

λ

λ

( )

( )( )

(7.16)

K10869_Book.indb 100 8/26/09 2:01:13 PM


wherem is total number of secondary resources.Ci is cost associated with secondary resource i for i = 1, 2, 3,…, m.l is ratio of software secondary development cost to software primary

development cost.

Model II is concerned with estimating the duration of a software project. The model predicts the minimum project duration under the assumption that the total hardware will be available during the project life. Thus, the minimum project duration is expressed by [19]

D

D

Sp

amin = (7.17)

whereDmin is minimum project duration.Dp is total programmer-months.Sa is average staff size allocated to the software project under consideration.

Additional information on the model is available in Schneider [19].Model III is concerned with estimating the software marketing cost, which

is expressed by [20]

C C Csm ho fs= + (7.18)

whereCsm is annual software marketing cost.Cho is cost associated with the home office.Cfs is field sales-related cost.

The field sales-related cost is given by

C BS SAS r r APSfs o c= + + +[ ( ) ( ) ]( ) ( )θ θ α 1 (7.19)

whereq is total number of people involved in sales.BS is annual base salary of a salesperson.SAS is annual salary of a system analyst.a is total number of system analysts employed per salesperson.ro is overhead rate.rc is commission rate.APS is annual product sales.

Model IV is concerned with estimating the software quality cost, which is expressed by [21]

SQC PC AC IFC EFC= + + + (7.20)

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whereSQC is software quality cost.PC is prevention cost associated with activities performed to prevent the

occurrence of software errors. Some examples of these activities are developing a software quality infrastructure, improving and updat-ing that infrastructure, and performing the regular activities neces-sary for its successful operation.

AC is appraisal cost associated with activities pertaining to the detection of software errors in software projects under consideration. Some examples of the appraisal cost components are the cost of software testing, cost of reviews, and cost of assuring quality of external par-ticipants (e.g., subcontractors).

IFC is internal failure cost associated with correcting software errors dis-covered through testing, design reviews, and acceptance tests prior to the installation of the software under consideration at customer sites.

EFC is external failure cost associated with correcting software failures dis-covered by customers after the installation of software at their sites.

Model V is concerned with estimating software project effort, in program-mer-months, in situations when very little information about the project under consideration is available, except its expected delivery instructions. The software project effort is expressed by [19]

SPE I S Ld cf af= ( . )( )( )( )1 7 (7.21)

whereSPE is software project effort expressed in programmer-months.Id is delivered instructions expressed in thousands.Scf is software complexity factor. Its values for trivial, moderately complex,

and very complex software are 1, 5, and 10, respectively.Laf is labor estimate adjustment factor expressed in decimal fraction. Its

recommended values for rather poorly managed projects and under best conditions are 2.9 and 0.435, respectively.

Model VI was developed by the U.S. Naval Air Development Center and is concerned with estimating the effort to develop software [22]. This effort is expressed by [17,22]

SDE x y z K L M= + + + + − −2 8 1 3 33 10 17 188. . (7.22)

whereSDE is total number of man-months needed for the software development.x is delivered program’s machine language instructions expressed in

thousands.y is contractor man-miles traveled.z is total number of document types produced or generated.

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K is total number of independent consoles in the delivered system.L is number of new instructions in percentages.M is average programmer experience with the system under consideration,

expressed in years.

Problems

1. Write an essay on computer system life cycle costing. 2. Assume that the acquisition cost and expected useful life of a

computer system are $5,000 and 6 years, respectively. The com-puter system’s expected number of failures per million hours is 80 and its only ownership cost is the cost of corrective maintenance. Calculate the life cycle cost of the computer system, if the cost of each corrective maintenance call is $150 and the annual discount or interest rate is 6%.

3. Discuss the major difficulties faced in estimating software cost. 4. Discuss the factors that influence software life cycle cost. 5. What is the difference between computer hardware and software

costing? 6. Write an equation that can be used to estimate software life cycle

cost. 7. Discuss software cost estimation methods. 8. Compare tabular models with linear models with respect to soft-

ware costing. 9. Discuss a mathematical model that can be used to estimate com-

puter system hardware maintenance cost. 10. What is the main difference between Equation (7.2) and Equation

(7.3)?

References

1. Carter faces problems in achieving his 1980 budget goals. 1979. Wall Street Journal, January 23: 4–5.

2. Keene, S. J. 1992. Software reliability concepts. Annual Reliability and Maintainability Symposium Tutorial Notes 1–21.


4. Phister, M. 1978. Analyzing computer technology costs—Part II: Maintenance. Computer Design October: 109–118.

5. Stanley, M. 1982. Software cost estimating, royal signals and radar establish-ment. Memorandum no. 3472. Procurement Executive, Ministry of Defense, Malvern, Worcs., UK.

K10869_Book.indb 103 8/26/09 2:01:45 PM


6. Boehm, B. W. 1984. Software life cycle factors. In Handbook of software engineer-ing, ed. C. R. Vick and C. V. Ramamoorthy, 494–518. New York: Van Nostrand Reinhold Company.

7. Dhillon, B. S. 1987. Reliability in computer system design. Norwood, NJ: Ablex Publishing Corporation.


9. Boehm, B. W. 1981. Software engineering economics. Englewood Cliffs, NJ: Prentice Hall, Inc.

10. Black, R. K. D., Curnow, R. P., Katz, R., and Gray, M. D. 1977. BCS software production data. Report no. RADC-TR-77-116. Boeing Computer Services, Inc. Available from the National Technical Information Services (NTIS), Springfield, VA.

11. Aron, J. D. 1969. Estimating resources for large programming systems, NATO Science Committee, Rome.

12. Wolverton, R. W. 1974. The cost of developing large-scale software. IEEE Transactions on Computers 23: 615–636.

13. Putnam, L. H. 1978. A general empirical solution to the macro software sizing and estimating problem. IEEE Transactions on Software Engineering 4: 345–361.

14. Halstead, M. H. 1977. Elements of software science. New York: Elsevier. 15. Walston, C. E., and Felix, C. P. 1977. A method of programming measurement

and estimation. IBM Systems Journal 16:54–73. 16. Herd, J. R., Postak, J. N., Russell, W. E., and Stewart, K. R. 1977. Software cost

estimation study—Study results. Report no. RADC-TR-77-220, vol. 1. Doty Associates, Inc., Rockville, MD.

17. James, T. G. 1977. Software cost estimating methodology. Proceedings of the IEEE National Aerospace and Electronics Conference 22–28.

18. Doty, D. L., Nelson, P. J., and Stewart, K. R. 1977. Software cost estimation study, vol. II. Report no. RADC-TR-77-220. Prepared by Doty Associates, Inc., Rockville, MD.

19. Schneider, V. 1978. Prediction of software effort and project duration—Four new formulas. Sigplan Notices 13:49–59.

20. Phister, M. 1976. Data processing technology and economics. Santa Monica, CA: Santa Monica Publishing Company.

21. Galin, D. 2004. Software quality assurance. Harlow, Essex, England: Pearson Education Ltd.

22. Buck, F. et al. 1971. A cost by function model for avionic computer systems. Report no. NADC-SD-7088, vol. 1. Developed by Naval Air Development Center, Warminster, PA.

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105

8Transportation System Life Cycle Costing

8.1 Introduction

Each year a vast amount of money is spent to develop, manufacture, and operate transportation systems such as motor vehicles, trains, aircraft, and ships throughout the world. This amount has become an important element of the global economy. Saving a small percentage of this amount can result in a large sum of money.

The concept of life cycle costing is increasingly being applied to make vari-ous types of decisions concerning transportation systems, particularly at their design and procurement stages. The main reason for the increasing use of the life cycle costing concept during a transportation systems’ design and procure-ment stages is that past experiences indicate that many transportation systems’ ownership costs (i.e., logistics and operating cost) often exceed their procure-ment costs. This is also the case for many other engineering products and sys-tems. In fact, according to Ryan [1], the ownership costs of certain engineering products and systems can vary from 10 to 100 times their acquisition costs.

Over the years, a large number of publications have appeared on various aspects of transportation system life cycle costing. This chapter presents various important aspects of aircraft, ship, urban rail, and motor vehicle life cycle costing.

8.2 Aircraft Life Cycle Cost

Although the life cycle cost breakdown structure of an aircraft can vary from one organization to another and from one type of aircraft to another, it can be broken down into four parts as follows:

LCC C C C Ca = + + +1 2 3 4 (8.1)

whereLCCa is aircraft life cycle cost.C1 is aircraft research, development, test, and evaluation cost.C2 is aircraft production cost.C3 is aircraft initial support costs associated with items such as support

equipment, spares, data, special equipment, and contractual training.

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C4 is aircraft operations and support cost associated with items such as base-level maintenance, training, and operations personnel and depot-level engine and component repair.

Usually, the life cycle cost of a typical fighter aircraft is broken into three categories [1]:

LCC DC PC OSCfa fa fa fa= + + (8.2)

whereLCCfa is life cycle cost of a typical fighter aircraft.DCfa is fighter aircraft development cost.PCfa is fighter aircraft acquisition cost.OSCfa is fighter aircraft operations and support cost.

According to Huie and Harris [2], for a typical fighter aircraft, the operations and support cost, acquisition cost, and development cost over its life span of 15 years usually account for approximately 55, 35, and 10% of the life cycle cost, respectively.

The four main components of fighter aircraft development cost are design and development cost, test and evaluation cost, flight test support cost (e.g., cost of spares, ground support equipment, and personnel), and cost of data (e.g., test and stress reports). Normally, activities such as design, manufactur-ing, and testing account for roughly 90% of the development cost. The factors that drive the cost of the development include mission capabilities, physical characteristics such as weight, size, reliability, and maintainability charac-teristics (e.g., mean time between failures and mean time to repair).

The six main components of the fighter aircraft acquisition cost are shown in Figure 8.1 [2,3]. Two of these components (i.e., flyaway cost and cost of initial support) account for an extremely large percentage of the acquisition cost. The flyaway cost includes the cost of the airframe, engine, and avionics, and the cost of initial support includes the cost of spares, ground support equipment, inventory entry and management, and training and training equipment.

Some of the main drivers of the acquisition cost are reliability and main-tainability characteristics, maintenance concept, mission capabilities, and training system requirements.

The fighter aircraft operations and support cost is composed of nine main components:

cost of fuel;•cost of personnel;•cost of depot maintenance;•cost of facilities;•cost of base maintenance material;•cost of modifications;•

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Transportation System Life Cycle Costing 107

cost of replenishing spares;•cost of replacement training; and•cost of item management.•

The five cost components that account for approximately 85% of the opera-tions and support cost are fuel cost, depot maintenance cost, personnel cost, base maintenance material cost, and replenishing spares cost.

The seven factors that drive the fighter aircraft operations and support cost are shown in Figure 8.2 [3].

The flyaway cost

The cost of data

The cost of facilities

The cost of initial support

The cost of system project management

Test and evaluation cost

Main components

FIGURE 8.1Main components of fighter aircraft acquisition cost.

Fuelconsumption

rate

Mean flight hours

betweenfailures

Unit/equipment

cost

Overhaulintervals

Maintenanceconcept

Utilizationrate

Force size Cost driving

factors

FIGURE 8.2Fighter aircraft operation and support cost driving factors.

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8.3 Aircraft Turbine Engine Life Cycle Cost

In the overall cost of an aircraft, the turbine engine is an important subsys-tem. The engine life cycle cost is expressed by [4,5]

LCC TEDC TEPIC TEPQC TEBMC TEDMCae = + + + + (8.3)

whereLCCae is aircraft turbine engine life cycle cost.TEDC is turbine engine development cost.TEPIC is turbine engine part improvement cost.TEPQC is turbine engine production quantity cost.TEBMC is turbine engine base maintenance cost.TEDMC is turbine engine depot maintenance cost.

Equations to estimate the five right-hand-side elements of Equation (8.3) are given in Jones [4] and Nelson [5].

8.4 Aircraft Cost Drivers

There are many aircraft cost drivers. In general, they may be grouped under the following three areas [2]:

design;•manufacturing; and•operations and support.•

The design cost drivers may be divided into three categories: reliabil-ity and maintainability requirements, performance requirements, and specifications. Two important elements of the reliability and maintainability requirements are mean time between failures (MTBF) and mean time to repair (MTTR).

Similarly, the four elements of the performance requirements are speed, payloads, range, and mission role. Finally, the elements of the specifications include corrosion control and fatigue life. Four main categories of the typical manufacturing cost drivers are shown in Figure 8.3 [2,3].

The elements of the material category include steel, aluminum, titanium, and composite. Some of the elements of the manufacturing process category are forgings, castings, machined parts, and sheet metal. The two main ele-ments of the structure category are wing and body. The wing includes com-ponents such as the number of hard points, wet versus dry, and complexity of control surfaces. Similarly, the body includes components such as landing gear attachment and wing attachment.

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The subsystems category has two main elements: flight control and land-ing gear. The flight control includes items such as the number of redundan-cies and mechanical versus fly-by-wire. Similarly, the landing gear includes items such as the number of wheels and brakes.

8.4.1 Helicopter Maintenance Cost Drivers

Many maintenance cost drivers are associated with helicopters. For exam-ple, maintenance cost drivers for military helicopters include the rotor sys-tem, power plants, transmissions, inspections, and others [3]. Generally, the breakdown percentages of direct maintenance cost (parts and labor) for these cost drivers are roughly 29, 27, 12, 9, and 23%, respectively. The breakdown percentages within the rotor system are blades (80%) and hub (20%).

Furthermore, note that the major contributor to the rotor hub operation and support cost is the seal leak, which results in lubricant loss and fluid. Similarly, two major contributors to the rotor blade operation and support cost are foreign object damage and inability to repair damaged blades.

8.4.2 Aircraft Airframe Maintenance Cost Drivers

According to a study performed in the early 1970s, the top nine airframe maintenance cost components were as follows [3]:

brakes;•tires;•

Structure (wing and body) Manufacturing

process

MaterialSubsystems

(landing gear and flight control)

Cost driver categories

FIGURE 8.3Main categories of typical aircraft manufacturing cost drivers.

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nose landing gear wheel and tire;•flight control power control units;•constant speed drive;•motor-driven hydraulic pump;•engine-driven hydraulic pump;•auxiliary power unit; and•starter.•

8.4.3 Combat Aircraft Hydraulic and Fuel Systems Cost Drivers

The following are cost drivers of a combat aircraft hydraulic system [3,6]:

valves (9• %);pumps (26• %);filters (12• %);reservoirs (20• %);accumulators (7• %);plumbing (12• %); andother (14• %).

The cost drivers of a combat aircraft fuel system are valves (33%), pumps (27%), filters (8%), measurement (17%), and other (15%) [3,6].

8.5 Cargo Ship Life Cycle Cost

Ships are an important mode of transportation; over 90% of the world’s cargo is transported by merchant ships. The life cycle cost of a cargo ship is expressed by [7]

LCC BC OC AC OPCCS = + + + (8.4)

whereLCCCS is cargo ship life cycle cost.BC is cargo ship building cost, including the cost of items such as machin-

ery, outfitting, and hull.OC is the cargo ship owner’s cost, including items such as naval architect’s

fee, attorney’s fee, and consulting fees.AC is cargo ship accommodation cost, including the cost of items such as

steel, hull engineering, and outfiting.

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OPC is cargo ship operating cost, including the cost of items such as fuel, maintenance and repair, cargo handling, part changes, wages, inven-tory, protection and indemnity insurance, hull and machinery insur-ance, and subsistence.

Additional information on cargo ship life cycle cost is available in Earles [7].

8.6 Operating and Support Costs for Ships

Over the years, many formulas have been developed to estimate various types of ship operating and support costs. Some of these formulas that have been developed for the U.S. Navy are presented next [7,8].

8.6.1 Formula I

This formula is concerned with estimating the cost of repair parts:

C A B LDrp = + ( )( ) (8.5)

whereCrp is cost of repair parts expressed per steaming hour (i.e., underway and

not underway) in 1976 dollars.A = 28.083B = 0.00263LD is full load displacement expressed in tons.

8.6.2 Formula II

This formula is concerned with estimating the cost of conventional fuel and is expressed by

C D E HP F xcf = + −( )( ) ( ) (8.6)

whereCcf is cost of conventional fuel.D = 166.021E = 0.001974HP is total shaft horsepower.F = 490.220x is a dummy variable whose value is either 1 (when the ship is nuclear

powered) or 0 (when the ship is not nuclear powered).

8.6.3 Formula III

This formula is concerned with estimating the ship overhaul cost, which is expressed by

C MD N MD NSho r r= +( )( ) ( . )( )( )0 25 (8.7)

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whereCSho is ship overhaul cost.MDr is repair man-days per overhaul.N = $150 (in 1976 dollars)

Note that the right-hand side of Equation (8.7) is composed of two compo-nents: labor cost [(MDr) (N)] and material cost [(0.25) (MDr) (N)].

8.6.4 Formula IV

This formula is concerned with estimating ship supplies’ cost. This is expressed by

C G H I xSS = + +( ) ( ) ( )α (8.8)

whereCSS is ship supplies’ cost.G = 44,797.515H = 248.260a is ship crew size (i.e., officers + enlisted individuals).I = 478,830x is a dummy variable whose value is either 1 (i.e., when the ship is nuclear

powered) or 0 (i.e., when the ship is not nuclear powered).

Note that Equation (8.8) provides the annual cost of health, safety, and wel-fare supplies expressed in 1976 dollars.

8.7 Urban Rail Life Cycle Cost

Urban rail is an important means of transportation around the globe. Each day it transports millions of passengers and millions of dollars worth of goods from one place to another. The urban rail life cycle cost is defined by [7]

LCC SCC SOCur = + (8.9)where

LCCur is life cycle cost.SCC is capital cost, including the cost of items such as vehicles, track and

track work, power substations and distribution, stations, and yard and maintenance facilities.

SOC is operating cost, including the cost of items such as power, trans-portation-associated manpower, and maintenance of tracks, vehicles, and equipment.

Additional information on this topic is available in references 3, 7, and 9.

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8.8 Car Life Cycle Cost

Each day, a vast sum of money is spent to procure and operate various types of cars throughout the world. The life cycle cost of a car is defined by [3,10]

LCC C OC SMC USMC CC a j j j dj

n

= + + + +=

∑1

(8.10)

whereLCCC is life cycle cost of the car.Ca is acquisition cost.n is expected life of the car expressed in years.OCj is operating cost (i.e., for gas, oil, tires, etc.) for year j for j = 1, 2, 3,…, n.SMCj is scheduled maintenance cost (i.e., for tune-up, lubrication, etc.) for

year j for j = 1, 2, 3,…, n.USMCj is unscheduled maintenance or repair cost (dependent on car fail-

ure rate) for year j for j = 1, 2, 3,…, n.Cd is car disposal plus any other cost.

Additional information on car life cycle costing is available in references 3, 7, and 10.

Example 8.1Assume that the acquisition cost of a car is $23,000. Annual scheduled and unscheduled maintenance costs are $200 and $400, respectively. Furthermore, the annual operating cost of the car is $1,500 and its life expectancy is 7 years. Calculate the car life cycle cost, if its disposal cost and the annual interest rate are $2,000 and 4%, respectively.

By using an equation given in Chapter 2 and in reference 3 and the given data, we get the following present values of the car operating cost, scheduled mainte-nance cost, and unscheduled maintenance cost, respectively:

OCp =

− +

=

−1500

1 1 0 040 04

9 003 17( . )

.$ , .

SMCp =

− +

=

−200

1 1 0 040 04

1 200 47( . )

.$ , .

and

USMCp =

− +

=

−400

1 1 0 040 04

2400 87( . )

.$ .

whereOCp is present value of operating cost.SMCp is present value of scheduled maintenance cost.USMCp is present value of unscheduled maintenance cost.

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Using an equation given in Chapter 2 and in reference 3 and the specified data values, we get the following present value of the car disposal cost:

Cdp =

+=2 000

1 0 041 519 8

7

,( . )

$ , .

where Cdp is present value of the car disposal cost.Using all of the preceding calculated values, the given data value, and Equation

(8.10), we get the following value for the car life cycle cost:

LCCC = + + + +$ , $ , . $ , . , . $ , .23 000 9 003 1 1 200 4 2 400 8 1 519 8

== $ , .37 124 1

Thus, the car life cycle cost is $37,124.10.

8.9 City Bus Life Cycle Cost Estimation Model

A bus is an important means of transport in cities throughout the world. This model is concerned with estimating city bus cost over the life span of the vehicle. Thus, the city bus life cycle cost is expressed by [3,11]

LCC VAC TC IOC WC LC FC

MCC RC GOC OHC CIC TC

Cb = + + + + +

+ + + + + + (8.11)

whereLCCCb is city bus life cycle cost.VAC is vehicle acquisition cost.TC is tire cost.IOC is cost of intermediate overhauls.WC is cost of wages.LC is lubricant cost.FC is fuel cost.MCC is cost of maintenance and checkup.RC is repair cost.GOC is cost of general overhauls.OHC is cost of overhead.CIC is cost of compulsory insurance.TC is taxes.

Additional information on the model is available in Zalud and Lanc [11].

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Problems

1. Write an essay on transportation system life cycle costing. 2. What are the main components of fighter aircraft development

cost? 3. What are the driving factors of fighter aircraft operation and sup-

port cost? 4. What are the main components of the fighter aircraft procurement

cost? 5. Mathematically, define the life cycle cost of an aircraft turbine

engine. 6. Discuss helicopter maintenance cost drivers. 7. What are the aircraft airframe maintenance cost drivers? 8. Mathematically, define the life cycle cost of a cargo ship. 9. Write formulas to estimate ship overhaul cost and conventional fuel

cost. 10. Assume that the procurement cost of a car is $30,000. The annual

scheduled and unscheduled maintenance costs are $300 and $500, respectively. Furthermore, the annual operating cost of the car is $1,000 and its life expectancy is 8 years. Calculate the car life cycle cost, if its disposal cost and the annual interest rate are $1,500 and 6%, respectively.

References


2. Huie, E., and Harris, H. F. 1980. Balanced design—-minimum cost solution. In Design to cost and life cycle cost. North Atlantic Treaty Organization (NATO) Advisory Group for Aerospace Research and Development (AGARD) confer-ence proceedings no. 289, 13.1–13.14.


4. Jones, E. J. 1980. Design to life cycle cost interaction of engine and aircraft. In Application of design to cost and life cycle cost to aircraft engines. North Atlantic Treaty Organization (NATO) Advisory Group for Aerospace Research and Development (AGARD) lecture series no. 107, 3.1–3.15.

5. Nelson, J. R. 1980. An approach to the life cycle analysis of aircraft turbine engines. In Application of design to cost and life cycle cost to aircraft engines. North Atlantic Treaty Organization (NATO) Advisory Group for Aerospace Research and Development (AGARD) lecture series no. 107, 2.1–2.27.

6. Grieser, H. 1980. Impact on system design of cost analysis of specifications and requirements. In Design to cost and life cycle cost. North Atlantic Treaty Organization

K10869_Book.indb 115 8/26/09 2:03:07 PM


(NATO) Advisory Group for Aerospace Research and Development (AGARD) con-ference proceedings no. 289, 6.1–6.10.


8. Eskew, H. L, Frazier, T. P., and Heilig, P. T. 1977. An operating and support cost model for aircraft carriers and surface combatants. Report no. ADA044744. Administrative Science Corporation, Alexandria, VA. Available from National Technical Information Service (NTIS), Springfield, VA.

9. Griffin, T. 2007. Impact assessment of a possible urban rail initiative. Report no. ITLR-T17297-005. Prepared by Interfleet Technology Ltd., Pride Parkway, Derby, UK.

10. Bhuyan, S. K. 1982. Cost of quality as a customer perception. Proceedings of the American Society for Quality Control Annual Congress 459–464.

11. Zalud, F. H., and Lanc, J. 1972. Automobile reliability: A key to lower overall transport costs. Proceedings of the 14th International Automobile Technical Congress of FISITA, London, 6125–6132. Published by the Institution of Mechanical Engineers, London.

K10869_Book.indb 116 8/26/09 2:03:07 PM

117

9Civil Engineering Structures and Energy Systems Life Cycle Costing

9.1 Introduction

In recent years, energy conservation has received considerable attention because the escalation of fuel prices has made energy costs an important consideration in the procurement of a wide range of items or systems. In the development and construction of many civil engineering systems and build-ings, cost has become an increasingly important issue because past experi-ence indicates that operating and maintenance costs over the long life of a system or building far exceed initial costs. Thus, operating and maintenance costs must be factored into the decision process concerning procurement and construction of civil engineering systems and buildings because it may be more cost effective to take on a higher initial cost in order to obtain lower ownership costs of these items.

The concept of life cycle costing has frequently been used in making pro-curement and construction decisions concerning energy and civil engineer-ing systems. Over the years, a large number of publications on both these areas have appeared. This chapter presents various important aspects of civil engineering and energy systems life cycle costing.

9.2 Building Life Cycle Cost

In the past, decisions in the building industrial sector during the design phase were made basically by comparing initial capital costs. The main rea-son for using this approach was its simplicity. Various studies conducted over the years indicate that a building’s long-term costs can far outweigh initial capital costs [1,2].

Thus, estimating the life cycle cost of a building at the initial design stage is very important, because past experiences indicate that the earliest decisions tend to establish boundaries to a certain degree for the later ones. According to Khanduri, Bedard, and Alkass [2], around 75–95% of the total life cycle costs of a typical building are locked in by the time its working drawings are pre-pared. Furthermore, if an estimate of the total life cycle cost is available at an

K10869_Book.indb 117 8/26/09 2:03:08 PM


early design stage of a building project, then it is relatively easy to take appro-priate cost reduction measures. However, once the project goes into construc-tion, chances to influence the total project cost are reduced quite significantly.

Building life cycle cost is defined by [2]

LCC CC OC RMC DCb = + + + (9.1)

whereLCCb is life cycle cost of a building.CC is capital cost, which is composed of land and construction costs.OC is operation cost associated with items such as energy, insurance,

and wages.RMC is repair and maintenance cost.DC is demolition cost.

9.3 Steel Structure Life Cycle Cost

Life cycle cost of a steel structure is the total cost during its life span. Mathematically, it is expressed as follows [3,4]:

LCC IC MC INC RC OC FC DCSt = + + + + + + (9.2)

whereLCCSt is life cycle cost of a steel structure.IC is initial cost. This includes cost of planning and design; erection cost;

cost of preparing the project site, including the cost of the founda-tion; storage, handling, and receiving costs associated with fabricated pieces and rolled sections; material cost of structural members such as columns, bracings, and beams; fabrication cost, including the mate-rial costs of connection elements or components; cost associated with operation of machinery and tools at the construction site; and cost associated with transporting rolled sections to the fabrication shop and transporting the fabricated pieces to the construction site [4].

MC is maintenance cost associated with items such as painting of exposed parts of the steel structure.

INC is inspection cost associated with preventing potentially severe dam-age to the structure.

RC is repair cost.OC is operating cost associated with the proper use of the structure for

items such as electricity and heating.FC is probable failure cost. This cost is based on an acceptable probability

of failure.DC is demolishing or dismantling cost.

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Civil Engineering Structures and Energy Systems Life Cycle Costing 119

Past experience indicates that the following main factors influence the life cycle cost of a steel structure [3]:

structure maintenance policy;•structure usage;•cost of the rolled sections used in initial structure construction;•project site’s geographic location;•expected life of the structure;•total number of different section types employed in the structure •under consideration;total number of connections;•structure importance;•perimeter of rolled sections in the complete structure;•currency discount rate; and•total weight of all rolled sections used in the entire structure.•

9.4 Bridge and Waste Treatment Facilities Life Cycle Costs

Life cycle cost analysis is a powerful tool that allows bridge owners or man-agers to consider the potential consequences of their decisions in present day monetary terms. The life cycle cost of a bridge is expressed by [5,6]

LCC CONC INSC DESC FAIC RAMCbr = + + + + (9.3)

whereLCCbr is bridge life cycle cost.CONC is construction cost.INSC is inspection cost.DESC is design cost.FAIC is failure cost.RAMC is repair and maintenance cost.

The life cycle cost of waste treatment facilities is defined by [7]

LCC CONC EDIC OPC DDC SRC WTDC FECω = + + + + + + (9.4)

whereLCCw is waste treatment facilities life cycle cost.CONC is construction cost, which contains the cost of items such as build-

ing construction, process equipment, construction management, improvements to land, and site work.

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EDIC is engineering, design, and inspection cost.OPC is operating cost, including the cost of items such as materials, staff,

maintenance, peripheral equipment, and utilities.DDC is decontamination and decommissioning cost. It includes the cost

of decontamination and decommissioning (DAD) as well as the cost associated with managing the wastes generated during DAD.

SRC is start-up and readiness review cost and includes the cost of items such as training of personnel, operation and maintenance manuals, initial system testing, and preparation for and performance of con-tractor readiness reviews.

WTDC is waste transport and disposal cost.FEC is front-end cost. Usually, this cost includes mostly the cost of activities

that are not directly related to producing a new facility but rather are related to regulation. The other important components of the front-end cost are project management cost and cost of preliminary studies such as establishing project definition and developing functional and operational requirements.

9.5 Building Energy Cost Estimation

Over the years, a number of formulas have been developed to estimate the cost of various items concerned with building energy. Some of these formu-las are presented next [8,9].

9.5.1 Formula I

This formula is concerned with estimating annual lighting cost and is expressed by [8]

LC

BS OT ECa = ( )( )( )

1000 (9.5)

whereLCa is annual lighting cost.BS is light bulb size.OT is light bulb operating period.EC is electricity cost expressed in dollars per kilowatt hour.

Example 9.1Assume that a 100 W incandescent light bulb is operated for 9 hours per day for 365 days. Calculate the cost to operte the bulb during the specified period if the electricity cost is $0.4 per kilowatt hour.

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LCa = ×

=

( )( )( . )

$ .

100 9 365 0 41000

131 4

Thus, the total cost to operate the bulb during the given period will be $131.40.

9.5.2 Formula II

This formula is concerned with estimating annual water heating cost and is expressed as follows [8]:

WHC

BTU FCER Btua

a

p

=( )( )

( ) (9.6)

whereWHCa is annual water heating cost.BTUa is annual British thermal units.FC is cost per fuel unit.ER is efficiency (i.e., the ratio of energy output to energy input).Btup is British thermal unit per fuel unit.

Example 9.2Assume that, for a certain manufacturing process, 1,200 gallons of water per hour are needed and water temperature is at 170°F supplied at 60°F for 8 hours per day for 280 days per year. Furthermore, to heat the water, natural gas is burned at 70% efficiency level and its cost is $5 per 1,000 cubic feet (CF). Calculate the annual water heating cost.

By inserting the given data values into Equation (9.6), we get

WHCa = −[( ) ( . ) ( )( )( )]($ /1200 8 34 170 60 8 280 51000CCFBTU CF

)( . ) ( / )

$ , .

0 7 1000

17 614 08=

Thus, the total annual water heating cost will be $17,614.08.

9.5.3 Formula III

This formula is concerned with estimating air filter energy cost. The energy cost, Ce, over the useful life of the filter is expressed by [9,10]

C C A FL R K MBe p q f e= ( )( )( )( )( )/( )( , )10 000 (9.7)

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whereCp is power cost expressed in dollars per kilowatt hour.Aq is quantity of air to be filtered expressed in cubic feet per minute.FL is useful life of the air filter expressed in hours.Rf is filter final resistance expressed in inch water gauge.K is the constant with value 1.173. MBe is motor and blower efficiency.

9.5.4 Formula IV

This formula is concerned with estimating the cost of heat exchangers, and is expressed by [9,11]

C SAhen=θ ( ) (9.8)

whereChe is procurement cost, free-on-board (F.O.B) factory.SA is heat exchanger surface area expressed in square feet.q and n are constants (their tabulated values are available in references 9

and 11).

9.5.5 Formula V

This formula is concerned with estimating operational equipment energy consumption cost and is expressed by [12]

C P EP OH OEoe a C= ( )( )( )( ) (9.9)

whereCoe is total energy consumption cost of operational equipment.Pa is average electrical power rating.EPC is electrical power cost.OH is total number of annual operating hours.OE is total number of pieces of operational equipment.

9.6 Appliance Life Cycle Costing

There are many different types of appliances—for example, refrigerators, ranges and ovens, freezers, gas dryers, washing machines, electric dryers, and room air conditioners. Their life cycle costs can be estimated by using the following equation [9,13]:

LCC AQC ECFC fr

dra i

i

ii

K

= ++

+

=∑ ( )

( )1

11

(9.10)

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whereLCCa is appliance life cycle cost.AQC is appliance acquisition cost expressed in dollars.K is appliance useful life expressed in years.ECi is energy consumption of year i expressed in British thermal units

(BTUs).FC is annual fuel cost expressed in constant dollars per million BTUs.fr is annual fuel escalation rate (%) expressed in constant dollars.dr is discount rate (%) expressed in constant dollars.

In the case of yearly constant energy consumption, EC, and the fuel escala-tion rate, fr, over appliance useful life, Equation (9.10) simplifies to

LCC AQC EC FCfr

dra

i

ii

K

= +++

=∑( )( )

( )( )11

1

(9.11)

Past experience indicates that acquisition cost for items such as refrigerators, electric ranges, and room air conditioners accounts for roughly 41, 38, and 59% of their life cycle costs, respectively [12].

9.7 Energy System Life Cycle Cost Estimation Model

This model was developed by the Center for Building Technology of the National Bureau of Standards for the U.S. Department of Energy [13]. The model takes into consideration all relevant costs over time of a building facility’s design, materials, operation, systems, and components. More specifically, it includes items such as initial investment cost, operation and maintenance cost, future replacement cost, and salvage and resale value.

Thus, the energy system life cycle cost is expressed by [12,13]

LCC EC IC SV NFOMC NRC RCes pv pv pv pv pv pv= + + + + + (9.12)

whereLCCes is present value of the energy system life cycle cost.ECpv is present value of the energy cost.ICpv is present value of the investment cost.SVpv is present value of salvage.NFOMCpv is present value of the annually recurring nonfuel operation and

maintenance cost.NRCpv is present value of the nonrecurring nonfuel operation and mainte-

nance cost.

Additional information on this model is available in references 12 and 13.

K10869_Book.indb 123 8/26/09 2:04:19 PM


9.8 Motor, Pump, and Circuit-Breaker Life Cycle Costs

This section presents mathematical models to estimate life cycle cost of a motor, a pump, and a circuit breaker.

9.8.1 Motor Life Cycle Cost Estimation Model

This model is concerned with estimating the life cycle cost of an electric motor, which is expressed by [9,14]

LCC C Cm ma mo= + (9.13)

whereLCCm is motor life cycle cost.Cma is motor acquisition cost.Cmo is motor operating cost.

Note that in Equation (9.13), the motor maintenance cost is assumed negli-gible. Using Dhillon [9], the present value of the motor operating cost, Cmoj, for year j may be expressed as follows:

PV C

ij moj

j

=+

11

(9.14)

wherePVj is present value of the motor operating cost, Cmoj, for year j.i is interest rate.

If the motor operational life is m years, then the present value of the motor total operating cost is expressed by

C C

iC

iCmot mo mo mo=

+

++

++1 2

2

31

11

11

1 iiC

imom

m

+ ++

31

1L (9.15)

whereCmot is present value of the total operating cost of the motor.Cmoj is motor operating cost in year j for j = 1, 2, 3,…, m.

The yearly operating cost of the motor can be calculated by using the fol-lowing equation [8,14]:

C

OH MS CEFFmo

e=( )( . )( )( )0 746 (9.16)

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whereCmo is motor operating cost per year expressed in dollars.OH is annual motor operating hours.MS is motor size expressed in horsepower.Ce is cost of electricity expressed in dollars per kilowatt hour.EFF is motor efficiency.

Example 9.3Assume that a 30-horsepower electric motor is operated for 3,000 hours annually. The cost of electrical energy is $0.2 per kilowatt hour. Calculate the annual cost to operate the motor if motor efficiency is 95%.


Cmo =

=

( )( . )( )( . ).

$ , .

3000 0 746 30 0 20 95

14 137 74

Thus, the annual cost to operate the motor will be $14,137.74.

9.8.2 Pump Life Cycle Cost Estimation Model

This model is concerned with estimating the life cycle cost of a pump, which is expressed by [15,16]

LCC IC EC IAC DC DTC OC MRC ENCp = + + + + + + + (9.17)

whereLCCp is pump life cycle cost.IC is pump initial cost, including the cost of items such as pump, pipe,

system, and auxiliary.EC is pump energy cost associated with various aspects of pump sys-

tem operation.IAC is pump installation and commissioning cost, including the cost

of training.DC is pump decommissioning or disposal cost, which also includes the

cost associated with restoration of the local environment and dis-posal of auxiliary services.

DTC is pump downtime cost associated with the production losses.OC is pump operation cost, which is basically the labor cost of normal

pump system supervision.MRC is pump maintenance and repair cost.ENC is pump environmental cost associated with contamination from

pumped liquid.

Each of these eight costs is described in detail in reference 16.

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The pump energy cost, EC, may be calculated by using the following for-mula [8]:

EC

PS PHS AOP Ce

p m

=( )( )( )( )

( )5300 θ θ (9.18)

wherePS is pump size expressed in gallons per minute (GPM).PHS is pump head size expressed in feet.AOP is pump annual operational period expressed in hours.Ce is cost of electricity expressed in dollars per kilowatt hour.qp is pump efficiency.qm is motor efficiency.

Example 9.4Assume that an 800 GPM pump with a total head size of 10 feet is operated for 1,500 hours per year. The pump and motor efficiency are 70 and 90%, respec-tively. Calculate the annual cost to operate the pump if the cost of electricity is $0.3 per kilowatt hour.


EC =

=

( )( )( )( . )( )( . )( . )

$ , .

800 10 1500 0 35300 0 7 0 9

1 078 117

Thus, the annual cost to operate the pump will be $1,078.17.

9.8.3 Circuit-Breaker Life Cycle Cost Estimation Model

This model is concerned with estimating the life cycle cost of a high-voltage circuit breaker. This cost is expressed by [9,17]

LCC CFC CMC UCcb = + + (9.19)

whereLCCcb is life cycle cost of the high-voltage circuit breaker.CFC is high-voltage circuit breaker fixed cost.CMC is high-voltage circuit breaker maintenance cost.UC is cost associated with the unavailability of power transmission and

distribution systems.

Problems

1. Write an essay on building life cycle costing. 2. Write an equation for estimating life cycle cost of a building. 3. Write an equation that can be used to estimate life cost of a steel

structure.

K10869_Book.indb 126 8/26/09 2:05:05 PM


4. Write at least 10 factors that influence the life cycle cost of a steel structure.

5. Write an equation that can be used to estimate life cycle cost of a bridge.

6. Assume that a 60 W incandescent light bulb is operated for 6 hours per day for 365 days. Calculate the cost to operate the bulb during the specified period if the electricity cost is $0.3 per kilowatt hour.

7. Write formulas for estimating the cost of (1) heat exchangers, and (2) filter energy.

8. Assume that for a certain manufacturing process, 1,000 gallons of water per hour is required and water temperature is at 150°F sup-plied at 70°F for 6 hours per day for 250 days per year. Furthermore, to heat the water, natural gas is burned at 60% efficiency level and its cost is $6 per 1,000 CF. Calculate the annual water heating cost.

9. Write an equation that can be used to estimate life cycle costs of refrigerators and washing machines.

10. Assume that a 20-horsepower electric motor is operated for 2,000 hours annually. The cost of electrical energy is $0.3 per kilowatt hour. Calculate the annual cost to operate the motor if the motor efficiency is 90%.

References

1. Flanagan, R., Norman, G., Meadows, J., and Robinson, G. 1989. Life cycle costing, theory and practice. London: BSP Professional Books.

2. Khanduri, A. C., Bedard, C., and Alkass, S. 1983. Life cycle costing of office build-ings at the preliminary design stage. Proceedings of the 5th International Conference on Civil and Structural Engineering Computing 1–8.

3. Sarma, K. C., and Adeli, H. 2002. Life cycle cost optimization of steel structures. International Journal for Numerical Methods in Engineering 55:1451–1462.

4. Sarma, K. C., and Adeli, H. 2000. Fuzzy discrete multi-criteria cost optimization of steel structures. Journal of Structural Engineering (ASCE) 126 (11): 1339–1347.

5. Rafiq, M. I., Chryssanthopoulos, M., and Onoufriou, T. 2005. Comparison of bridge management strategies using life cycle cost analysis. Proceedings of the 5th International Conference on Bridge Management, Inspections, Maintenance, Assessment, and Repair 578–586.

6. Frangopol, D. M., Estes, A. C., Augusti, G., and Ciampoli, M. 1997. Optimal bridge management based on lifetime reliability and life cycle costs. Proceedings of the International Workshop on Optimal Performance of Civil Infrastructure Systems 98–115.

7. Sivill, T. E., Stoddard, D. N., Smith, T. H., and Roesener, W. S. 1993. Use of life cycle cost estimates in the evaluation of proposed waste-treatment facilities. Proceedings of the Technology and Programs for Radioactive Waste Management and Environmental Restoration Conference 1797–1801.

8. Brown, R. J., and Yanuck, R. R. 1980. Life cycle costing: A practical guide for energy managers. Atlanta, GA: Fairmont Press.

K10869_Book.indb 127 8/26/09 2:05:05 PM


9. Dhillon, B. S. 1989. Life cycle costing: Techniques, models and applications. New York: Gordon and Breach Science Publishers.

10. Avery, A. H. 1977. Life cycle costing of high-efficiency air filters. Plant Engineering September: 80–83.

11. Kumana, J. D. 1984. Cost update on specialty heat exchangers. Chemical Engineering June: 169–172.


13. National Bureau of Standards. 1980. Life cycle cost manual for the Federal Energy Management Program. National Bureau of Standards handbook 135. U.S. Department of Commerce, Washington, D.C.

14. Ganapathy, V. 1983. Life cycle costing applied to motor selection. Process Engineering July: 51–52.

15. De Boer, G., and Greidanus, D. 2006. Utilization of customized hydraulics to elongate pump life and lower cycle costs. Proceedings of the Institution of Mechanical Engineers 9th European Fluid Machinery Congress on Applying the Latest Technology to New and Existing Process Equipment 95–102.

16. Hydraulic Institute, Office of Industrial Technology. 2001. Pump life cycle costs: A guide to LCC analysis for pumping systems. Report no. DOE/GO-102001-1190, U.S. Department of Energy, Washington, D.C.

17. Heising, C. R. 1979. Reliability and maintenance data needed for high-voltage circuit breakers when making life cycle cost studies. Proceedings of the International Reliability Conference for the Electric Power Industry 103–108.

K10869_Book.indb 128 8/26/09 2:05:05 PM

129

10Miscellaneous Cost Estimation Models

10.1 Introduction

Over the years, a large number of cost estimation models have been devel-oped in diverse areas ranging from software engineering to telecommunica-tion engineering. A cost estimation model may be described simply as an approach, based on programmatic and technical parameters, to calculating costs under consideration. More specifically, some of the possible dimen-sions of a cost estimation model include the elements of cost, time, and cost breakdown structure.

Many desirable features are associated with a cost estimation model; in designing such a model, the main factors that should be considered are fea-sibility of data requirements; operation ease; cost to develop, operate, and alter; capability for sensitivity analyses; speed to set up, operate, and change; inclusiveness and authoritativeness; and tolerance of input errors [1,2].

There are various types of cost estimation models: capital cost estimation models, operation and maintenance cost estimation models, life cycle cost esti-mation models, and so on. This chapter presents a number of models that were not covered in previous chapters. They can also be used to estimate various types of costs for performing life cycle cost analysis directly or indirectly.

10.2 Plant Cost Estimation Model

This model was developed by Cran to estimate plant cost in the chemical industry [3]. The total plant cost is expressed by [3,4]

TPC C C

C C Cd i

d d if

= +

= + ( )( )

(10.1)

whereTPC is total plant cost expressed in dollars.Cd is direct cost expressed in dollars.Ci is indirect cost expressed in dollarsCif is indirect cost factor.

K10869_Book.indb 129 8/26/09 2:05:11 PM


The direct cost, Cd, is defined by

C C C C C

d is di e dp= +( )( ) ( )( )

(10.2)

whereCis is cost of instruments.Cdi is direct cost factor associated with instruments.Ce is cost of equipment.Cdp is direct cost factor associated with the plant.

The following are the mean values for plant direct cost and instrument direct cost factors:

2.16 (for • Cdp)2.50 (for • Cdi)

The value of the indirect cost factor, Cif, can be estimated by using the fol-lowing equation:

C l nC

if d= −1 36 0 073. ( . )

(10.3)

Additional information on the model is available in Cran [3] and Ward [4].

10.3 Reliability Acquisition Cost Estimation Model

This model can be used to estimate reliability acquisition cost when state-of-the-art system acquisition cost and reliability improvement ratio compared to state of the art are known. The reliability acquisition cost is expressed by [5]

C AC l nra sa

= ( . ) ( )0 2 α (10.4)

whereCra is reliability acquisition cost.ACsa is state-of-the-art system acquisition cost.a is reliability improvement ratio compared to state of the art.

Additional information on the model is available in Winlund [5].

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Miscellaneous Cost Estimation Models 131

10.4 Development Cost Estimation Model

This model is concerned with estimating development cost by considering the reliability factor. Thus, the development cost is expressed by [6,7]

DC C Cr ir dr

= + (10.5)

whereDCr is development cost, considering the reliability.Cir is basic cost, independent of reliability.Cdr is cost, dependent on reliability (i.e., reliability-related cost).

The cost dependent on reliability, Cdr, is defined by

C C

MTBFMTBFdr s

i

s

=

θ

(10.6)

whereCs is “standard” cost to develop item, equipment, or system having “standard”

or current reliability.MTBFi is item, equipment, or system mean time between failures with

improved design.MTBFs is item, equipment, or system mean time between failures with

standard design.q is a constant whose value is to be estimated from empirical studies.

Let us now assume that the reliability of the standard design, Rs(t), and the reliability of the improved design, Ri(t), are respectively expressed by [8]

R t es

tMTBFS( ) =

−

(10.7)

and

R t ei

tMTBFi( ) =

−

(10.8)

wheret is time.Rs (t) is standard design reliability at time t.Ri (t) is improved design reliability at time t.

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By taking natural logarithms of Equation (10.7) and (10.8) and then rear-ranging them, we get, respectively,

MTBF

tl nR ts

s

= −

( )

(10.9)

and

MTBF

tl nR ti

i

= −

( )

(10.10)

By substituting Equations (10.9) and (10.10) into Equation (10.6) and then substituting the resulting equation into Equation (10.5), we get

DC C C

l nR tl nR tr ir s

s

i

= +

( )( )

θ

(10.11)

Note that the preceding equation makes use of time-dependent reliabili-ties of standard and improved item, equipment, or system designs instead of mean time between failures (i.e., MTBFs and MTBFi) as in the case of Equation (10.6). Additional information on the model is available in Hevesh [6] and Carhart and Herd [7].

10.5 Program Error Cost Estimation Model

This model is concerned with estimating the cost of program errors in a pro-gram and is expressed by [8]

PEC C Coj cj

j

m

= +=

∑ ( )1

(10.12)

wherePEC is total cost of errors in a program.m is total number of errors in a program.Coj is cost associated with the occurrence of error j.Ccj is cost associated with correcting error j.

Note that the cost elements associated with the error occurrence cost are lost equipment time cost, wasted manpower hours cost, etc. Similarly, the cost of cor-recting the error includes components such as equipment cost, supply cost, and manpower cost. Additional information on the model is available in Sontz [8].

K10869_Book.indb 132 8/26/09 2:06:15 PM


10.6 Cooling Tower Cost Estimation Model

This model is concerned with estimating cooling tower cost using operating data. This cost is expressed by [9]

C

LZ X Rt

=− +( )( ) ( . )( )586 39 2

(10.13)

Z

Twb

=− +

2790 0335 85 11 143( . )( ) .

(10.14)

R T Thw cw

= − (10.15)

X T Tcw wb

= − (10.16)

whereCt is cost of a cooling tower expressed in dollars.L is total heat load expressed in BTUs per hour.X is the temperature approach.R is cooling range.Twb is wet bulb temperature expressed in degrees Fahrenheit.Thw is hot water temperature expressed in degrees Fahrenheit.Tcw is cooled water temperature expressed in degrees Fahrenheit.

Additional information on the model is available in Zanker [9].

Example 10.1Calculate the cost of a cooling tower, if the following data values are given:

Thw = 120°F;Tcw = 80°F;Twb = 60°F; andL = 300 million BTUs per hour.

By substituting the given data values into Equations (10.13)–(10.16), we get

R

X

Z

= − == − =

=−

120 80 40

80 60 20

2790 0335 85 6

o

o

F

F

( . ) ( 00 1119 89

300 000 000119 89 20

1 143).

, ,( . ) ( )

. +=

=Ct −− +

=586 39 2 40

88 760 64( . ) ( )

$ , .

Thus, the cooling tower cost is $88,760.64.

K10869_Book.indb 133 8/26/09 2:06:39 PM


10.7 Storage Tank Cost Estimation Model

This model is concerned with estimating the cost of storage tanks and is expressed by [10]

C Cst b

= ( )λ (10.17)

whereCst is cost of a storage tank.Cb is base cost, in carbon steel, expressed in dollars.l is the material-of-construction factor.

The base cost in carbon steel for field-erected tanks (cone roofs and flat bottoms) is expressed by

C l nV l nV

b= − + exp ( )θ θ θ

1 2 32

(10.18)

whereCb is base cost in carbon steel for field-erected tanks.q j is the jth constant for j = 1 (q1 = 9.369), j = 2 (q2 = 0.1045), and j =

3 (q3 = 0.045355).V is tank volume in cubic meters (80 m3 ≤ V ≤ 45,000 m3).

Similarly, the base cost in carbon steel for shop-fabricated tanks (cone roofs and flat bottoms) is expressed by

C l nV l nV

b= + − exp ( )α α α

1 2 32

(10.19)

whereCb is base cost in carbon steel for shop-fabricated tanks.aj is the jth constant for j = 1 (a1 = 7.994), j = 2 (a2 = 0.6637), and j =

3 (a 3 = 0.063088).V is tank volume in cubic meters (5 m3 ≤ V ≤ 80 m3)

The values of l for construction materials such as stainless steel 304, stain-less steel 316, stainless steel 347, aluminum, copper, nickel, titanium, and monel are 2.4, 2.7, 3, 2.7, 2.3, 3.5, 11.0, and 3.3, respectively. Additional infor-mation on the model is available in Corripio, Chrien, and Evans [10].

10.8 Pressure Vessel Cost Estimation Model

This model is concerned with estimating the cost of pressure vessels. The total cost is expressed by [11]

PVC C Cp

= +( )θ υ (10.20)

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wherePVC is total cost of a pressure vessel expressed in dollars.q is construction material cost factor.Cu is base vessel cost in carbon steel expressed in dollars. Cp is cost of platform and ladders expressed in dollars.

For horizontal vessels, Cu and Cp are expressed by Equations (10.21) and (10.22), respectively:

C L L l n W L l nWυ = − + exp ( )

1 2 32

(10.21)

whereLj is the jth constant for j = 1 (L1 = 8.114), j = 2 (L2 = 0.16449), and j =

3 (L3 = 0.04333).W is carbon steel shell weight in kilograms (369 kg ≤ W ≤ 415,000 kg).

C M D

pM=

12( )

(10.22)

whereMj is the jth constant for j = 1 (M1 = 1288.3) and j = 2 (M2 = 0.20294).D is inside diameter of platform and ladders in meters (0.92 m ≤ D ≤ 3.66 m).

Similarly, for vertical vessels, Cu and Cp are expressed by Equations (10.23) and (10.24), respectively:

C N N l n W N l nWυ = − + exp ( )

1 2 32

(10.23)

whereNj is the jth constant for j = 1 (N1 = 8.6), j = 2 (N2 = 0.21651), and j = 3 (N3 =

0.04576).W is carbon steel shell weight in kilograms (2210 kg ≤ W ≤ 103,000 kg).

C n D TL

pn n=

12 3( )

(10.24)

wherenj is the jth constant for j = 1 (n1 = 1017), j = 2 (n2 = 0.73960), and j =

3 (n3 = 0.70684).D is inside diameter of platform and ladders in meters (1.83 m ≤ D ≤ 3.05 m).

The values of q for construction materials such as stainless steel 316, stain-less steel 304, titanium, nickel 200, monel 400, incoloy 825, and inconel 600 are 2.1, 1.7, 7.7, 5.4, 3.6, 3.7, and 3.9, respectively. Additional information on the model is available in Mulet, Corripio, and Evans. [11].

K10869_Book.indb 135 8/26/09 2:07:30 PM


10.9 New Aircraft System Spares Cost Estimation Model

This model is concerned with estimating spares cost for the new aircraft system. The model uses the spares cost for an existing aircraft system and adjusts it by a comparison factor reflecting the differences in system cost, reliability, hardware complexity, and repairability. The new aircraft system’s spares cost is defined by [12]

C Cna ea

= γ (10.25)

whereCna is new aircraft system spares cost.q is the comparison factor expressed in terms of operational and support

parameters.Cea is existing aircraft system spares cost.

The comparison factor, q, is expressed by

θ β=

qMIMI

n

e

(10.26)

whereq is quantifier of the cost impact associated with a shift in the classifica-

tion of spares from “base repaired” to “depot repaired” or vice versa between the two aircraft systems under consideration. Note that the value of this quantifier is equal to unity when the change in the ratio of the two classifications is zero.

MIn is new aircraft system’s calculated (estimated) maintenance index expressed as maintenance man-hours per flying hour.

MIe is existing aircraft system’s established maintenance index expressed as maintenance man-hours per flying hour.

The symbol b in Equation (10.26) is expressed by

β =

+

+fCC

fCC

fCC

n

e

n

e

n

e1

1

12

2

23

3

33

(10.27)

whereCnj is jth segment of the “fly-away” cost for the new aircraft system for j = 1

(airframe), j = 2 (propulsion), and j = 3 (equipment).Cej is jth segment of the “fly-away” cost for the existing aircraft system for

j = 1 (airframe), j = 2 (propulsion), and j = 3 (equipment).

K10869_Book.indb 136 8/26/09 2:07:43 PM


fj is jth fraction of the total investment spares “lay-in” value calculated for existing systems for j = 1 (airframe related), j = 2 (propulsion related), and j = 3 (equipment related).

Additional information on the model is available in Tyszkiewicz [12].

10.10 Satellite Procurement Cost Estimation Model

This model is concerned with estimating the procurement cost of satellites in 1974 dollars. The satellite procurement cost is expressed by [2,13]

C W W

s s s=

−λ λ1

2

(10.28)

whereCs is satellite procurement cost.lj is the jth constant for j = 1 (l1 = 1,970,300) and j = 2 (l2 = 0.592).Ws is the satellite’s total weight.

Additional information on the model is available in Hadfield [13].

10.11 Single-Satellite System Launch Cost Estimation Model

This model is concerned with estimating the total launch cost of a single-satellite system (circular orbits). The launch cost is defined by [13]

LC W

OAf

s s p= + +

ββ

β1

3

2 1( )

(10.29)

whereLCs is launch cost of a single-satellite system, expressed in millions of

1974 dollars.bj is the jth constant for j = 1 (b1 = 0.026), j = 2 (b2 = 2/3), and j = 3 (b3 = 8,000).Ws is total satellite weight expressed in pounds.OA is orbit altitude or apogee expressed in statute miles.fp is a factor measuring the satellite payload sophistication.

Additional information on the model is available in Hadfield [13].

K10869_Book.indb 137 8/26/09 2:07:55 PM


10.12 Tank Gun System Life Cycle Cost Estimation Model

This model is concerned with estimating tank gun life cycle cost by decompos-ing it into three major components: research and development cost, investment cost, and operating and support cost. The life cycle cost is expressed by [1,2]

LCC RDC IC OS

tg tg tg tg= + +

(10.30)

whereLCCtg is tank gun system life cycle cost.RDCtg is tank gun system research and development cost.ICtg is tank gun system investment cost.OStg is tank gun system operating and support cost.

The tank gun system research and development cost, RDCtg, is expressed by

RDC RDCtg tgj

j

==

∑1

10

(10.31)

whereRDCtgj is cost component j of the tank gun system research and develop-

ment cost forj = 1 (tooling cost)j = 2 (facilities cost)j = 3 (development engineering cost)j = 4 (system project management cost)j = 5 (prototype manufacturing cost)j = 6 (system test and evaluation cost)j = 7 (training cost)j = 8 (producibility engineering and planning cost)j = 9 (data cost)j = 10 (other cost)

The tank gun system investment cost, ICtg, is expressed by

IC ICtg tgj

j

==

∑1

11

(10.32)

whereICtgj is cost component j of the tank gun system investment cost forj = 1 (training cost)j = 2 (production cost)

K10869_Book.indb 138 8/26/09 2:08:11 PM


j = 3 (data cost)j = 4 (nonrecurring investment cost)j = 5 (system project management cost)j = 6 (initial spares and repair parts cost)j = 7 (engineering changes cost)j = 8 (transportation cost)j = 9 (system test evaluation cost)j = 10 (operational and site activation cost)j = 11 (other cost)

The tank gun system operation and support cost, OStg, is expressed by

OS OStg tgj

j

==

∑1

6

(10.33)

whereOStgj is the cost component j of the tank gun system operating and support

cost forj = 1 (consumption cost)j = 2 (modification material cost)j = 3 (military personnel cost)j = 4 (depot maintenance cost)j = 5 (other direct support operations cost)j = 6 (indirect support and operations cost)

Additional information on the model is available in Earles [1] and Dhillon [2].

10.13 Weather Radar Life Cycle Cost Estimation Model

This model is concerned with estimating the life cycle cost of weather radar. This cost is expressed by [1]

WRLCC SDC VC AC OMSC= + + + (10.34)

whereWRLCC is weather radar life cycle cost.SDC is weather radar system definition cost.VC is weather radar validation cost.AC is weather radar acquisition cost.OMSC is weather radar operation, maintenance, and support cost.

K10869_Book.indb 139 8/26/09 2:08:21 PM


The weather radar system definition cost, SDC, is expressed by

SDC SDCi

i

==

∑1

2

(10.35)

where SDCi is the ith cost component of the weather radar system definition cost for i = 1 (program management cost) and i = 2 (cost for each bidder).

The weather radar system validation cost, VC, is expressed by

VC VCi

i

==

∑1

15

(10.36)

whereVCi is the ith cost component of the weather radar validation cost fori = 1 (engineering design and development cost)i = 2 (fabrication and manufacturing development cost)i = 3 (validation hardware cost)i = 4 (software system design and development cost)i = 5 (logistics planning and support cost)i = 6 (development test and test support cost)i = 7 (validation test site preparation cost)i = 8 (documentation cost)i = 9 (manual development cost)i = 10 (support and test equipment cost)i = 11 (training development and equipment cost)i = 12 (government-furnished equipment and facilities cost)i = 13 (transportation of equipment to test site cost)i = 14 (program management cost)i = 15 (general and administration cost)

The weather radar acquisition cost, AC, is expressed by

AC ACi

i

==

∑1

18

(10.37)

whereACi is ith cost component of the weather radar acquisition cost fori = 1 (software and firmware-manufacturing-related cost)i = 2 (software and firmware-depot-related cost)i = 3 (software and firmware-on-site-related costs)i = 4 (initial training cost)i = 5 (vendor warranty for first year cost)i = 6 (test and support equipment cost)i = 7 (initial spares cost)

K10869_Book.indb 140 8/26/09 2:08:40 PM


i = 8 (test and evaluation cost)i = 9 (data and documentation cost)i = 10 (site preparation cost)i = 11 (system installation and checkout cost)i = 12 (site decommissioning cost)i = 13 (land acquisition cost)i = 14 (government printing cost)i = 15 (manual binding and delivery cost)i = 16 (government-furnished equipment cost)i = 17 (program management cost)i = 18 (general and administration overhead cost)

The weather radar operation, maintenance, and support cost, OMSC, is expressed by

OMSC OMSCi

i

==

∑1

13

(10.38)

whereOMSCi is ith cost component of the weather radar operation, maintenance,

and support cost fori = 1 (operating personnel cost)i = 2 (electric power cost)i = 3 (communications facilities cost)i = 4 (occupying and housekeeping cost)i = 5 (consumables cost)i = 6 (dedicated maintenance personnel cost)i = 7 (other maintenance-preventive and corrective cost)i = 8 (recurring spares cost)i = 9 (logistics and logistics support cost)i = 10 (other maintenance-test and support cost)i = 11 (equipment rental and housekeeping cost)i = 12 (maintenance by contractor cost)i = 13 (recurring training cost)

Additional information on the model is available in Earles [1].

Problems

1. What is a cost estimation model? 2. Write an essay on cost estimation models. 3. Discuss the desirable features of a cost estimation model.

K10869_Book.indb 141 8/26/09 2:08:46 PM


4. Define two main elements of a plant cost estimation model. 5. Define the following two models: (1) reliability acquisition cost

estimation model, and (2) development cost estimation model. 6. Define program error cost estimation model. 7. If the following data values are known, estimate the cost of a cool-

ing tower by using Equation (10.13):L• = 400 million BTUs per hourT• hw = 130°FT• cw = 90°FT• wb = 65°F

8. Define satellite acquisition cost estimation model. 9. Discuss the following two items: (1) weather radar life cycle cost

and (2) tank gun system life cycle cost 10. Define the following two models: (1) storage tank cost estimation

model and (2) pressure vessel cost estimation model.

References



3. Cran, J. 1981. Improved factored method gives better preliminary cost estimates. Chemical Engineering April: 65–79.

4. Ward, T. J. 1986. Cost-estimating methods. In Design of Equipment, vol. 1: Plant design and cost estimating, ed. J. Beckman, 12–21. New York: American Institute of Chemical Engineers.

5. Winlund, E. S. 1965. Cost-effective analysis for optimal reliability and maintain-ability. Proceedings of the Annual National Symposium on Reliability and Quality Control 107–114.

6. Hevesh, A. H. 1969. Cost of reliability improvement. Proceedings of the Annual Symposium on Reliability 54–61.

7. Carhart, R. R., and Herd, G. R. 1957. A simple cost model for optimizing reli-ability. In Reliability of military electronic equipment, a report by Advisory Group on Reliability of Electronic Equipment (AGREE), Office of the Assistant Secretary of Defense (Research and Engineering), Department of Defense, Washington, D.C., 64–74.

8. Sontz, C. 1973. Quality assurance for the data processing industry. Proceedings of the Annual Reliability and Maintainability Symposium 136–148.

9. Zanker, A. 1972. Estimating cooling tower costs from operating data. Chemical Engineering June: 118–120.

10. Corripio, A. B., Chrien, K. S., and Evans, L. B. 1982. Estimate costs of heat exchangers and storage tanks via correlations. Chemical Engineering January: 125–127.

K10869_Book.indb 142 8/26/09 2:08:46 PM


11. Mulet, A., Corripio, A. B., and Evans, L. B. 1981. Estimate costs of pressure ves-sels via correlations. Chemical Engineering October: 145–150.

12. Tyszkiewicz, A. M. 1983. A comparative cost model for aircraft investment spares. Proceedings of the Annual Reliability and Maintainability Symposium 376–382.

13. Hadfield, B. B. 1974. Satellite-systems cost estimation. IEEE Transactions on Communications 22:1540–1547.

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145

11Introduction to Engineering Reliability and Maintainability

11.1 Introduction

Reliability may be described simply as the probability that an item or sys-tem will perform its mission satisfactorily for the desired period when used according to designed conditions. The history of the reliability field may be traced back to the early 1930s, when probability-related concepts were applied to various problems associated with electric power generation [1–4]. During World War II, Germans applied various basic reliability concepts to improve reliability of their rockets (i.e., V1 and V2 rockets). A detailed his-tory of the reliability field is available in Dhillon [5]. Today, reliability is a well-established discipline and has branched out into many specialized areas [5,6].

Maintainability may be described as the aspects of equipment or an item that improve repairability and serviceability, increase cost effectiveness of maintenance, and ensure that the equipment or item satisfies the require-ments for its intended application. The roots of the maintainability history may be traced back to 1901 in the Army Signal Corps contract for develop-ment of the Wright Brothers’ airplane, which stated that the aircraft “should be simple to operate and maintain.” However, in the modern context, the beginning of the maintainability discipline may be traced back to the period between World War II and the early 1950s [7,8]. During this period, the U.S. Department of Defense conducted various studies that indicated startling results concerning the state of reliability and maintainability of equipment used by the three services [8–10].

Needless to say, today reliability and maintainability are well-established disciplines and, over the years, a vast amount of literature on both the topics has appeared [11,12]. This chapter presents various fundamental aspects of reliability and maintainability considered useful for direct or indirect appli-cations in life cycle costing.

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11.2 Reliability and Maintainability Definitions

Some of the commonly used terms and definitions in the area of reliability and maintainability follow [13–16]:

Reliability• is the probability that an item will perform its assigned mission satisfactorily for the desired period when used according to specified conditions.Maintainability• is the probability that a failed item will be restored to its satisfactory working state within a stated total downtime, when maintenance activity is started according to specified conditions.Failure• is the inability of an item to perform its specified function within defined guidelines.Downtime• is the time during which the item or product is not in a condition to perform its stated mission or function.Availability• is the probability that an item or equipment will be avail-able for service when required.Redundancy• is the existence of more than one means for performing a stated function.Useful life• is the length of time an item or piece of equipment func-tions within an acceptable level of failure rate.Maintenance• is all scheduled and unscheduled actions necessary to keep an item or piece of equipment in a serviceable state or restor-ing it to serviceability. It includes items such as inspection, testing, repair, modification, and servicing.Mission time• is the time during which the item or piece of equipment is carrying out its stated mission.

11.3 Bathtub Hazard Rate Curve

The curve shown in Figure 11.1 is widely used to describe failure rate of vari-ous types of engineering items. As shown in the figure, the bathtub hazard rate curve is divided into three regions: region I (burn-in period), region II (useful life period), and region III (wear-out period).

During the burn-in period, hazard rate (i.e., time-dependent failure •rate) decreases with time t. Some of the main reasons for the occur-rence of failures in this region are inadequate quality control, poor processes, substandard materials and workmanship, poor manufac-turing methods, inadequate debugging, and human error.

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Introduction to Engineering Reliability and Maintainability 147

During the useful life period (region II), the hazard rate remains •constant with respect to time t. Some of the main reasons for the occurrence of failures in this region are higher random stress than expected, undetectable defects, human errors, low safety factors, abuse, and natural failures.Finally, during the wear-out period (region III), the hazard rate •increases with time t. The main causes for the occurrence of failures in this region include poor maintenance, wear due to aging, wrong overhaul practices, short designed-in life of the item under consider-ation, wear due to friction and corrosion, and creep.

11.4 General Reliability, Mean Time to Failure, and Hazard Rate Formulas

A number of general formulas are commonly used to perform reliability analysis. Three of these formulas are presented next.

11.4.1 General Formula for Reliability

This general formula is expressed by [17]

R t et dt

t

( )( )

=∫− λ0

(11.1)

(Region I) (Region II) (Region III)

Burn-inperiod Useful life period Wear-out period

Time dependentfailure rate (or hazard rate)

0 Time t

FIGURE 11.1Bathtub hazard rate curve.

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whereR(t) is reliability at time t.t is time.l(t) is time-dependent failure rate (i.e., hazard rate).

Example 11.1Assume that the hazard rate of an engineering system is given by

l (t) = l (11.2)

where l is engineering system constant failure rate. Obtain an expression for the engineering system reliability by using Equation (11.1).


R t e

e

tdt

t

( )==

− ∫

−

0 λ

λ

(11.3)

Thus, Equation (11.3) is the expression for the engineering system reliability.

11.4.2 General Formula for Mean Time to Failure

This general formula can be expressed in the three different ways that fol-low [10]:

MTTF R t dt=∞

∫ ( )0

(11.4)

or

MTTF R s

s=

→0

lim

( )

(11.5)

or

MTTF t f t dt=∞

∫ ( )0

(11.6)

wheref (t) is the failure or probability density function.s is the Laplace transform variable.R(s) is the Laplace transform of R(t).MTTF is mean time to failure.

Example 11.2Assume that the reliability of a piece of engineering equipment is expressed by

R t e t( )= −λ (11.7)

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wheret is time.l is engineering equipment failure rate.

Obtain an expression for the engineering equipment mean time to failure.By substituting Equation (11.7) into Equation (11.4), we get

MTTF e dtt=

=

−

∞

∫ λ

λ

0

1

(11.8)

Thus, Equation (11.8) is the expression for the engineering equipment mean time to failure.

11.4.3 General Formula for Hazard Rate

This general formula can be expressed in the following three ways [10]:

λ ( )

( )( )

tf t

f t dtt=

− ∫10

(11.9)

or

λ ( )

( )( )

tf tR t

=

(11.10)

or

λ( )

( ).

( )t

R td R t

dt= − 1

(11.11)

Example 11.3Using Equation (11.7), obtain a hazard rate expression for the engineering equip-ment. Comment on the resulting expression.


λ

λ

λ

λ( )t

ede

dtt

t

= −

=

−

−1

(11.12)

Thus, Equation (11.12) is the expression for the engineering equipment hazard rate. Note from this expression that the hazard rate is independent of time. Thus, it is simply referred to as the constant failure rate.

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11.5 Common Reliability Networks

Engineering systems can form various types of configurations or networks in performing reliability analysis. Some of the commonly occurring of these networks are presented next.

11.5.1 Series Network

This is the simplest and probably the most commonly occurring reliability network in engineering systems. Its block diagram is shown in Figure 11.2. Each block in the figure denotes a unit or component. More specifically, the Figure 11.2 diagram represents a system composed of m units in series. If any one of the units fails, the system fails. In other words, all system units must work normally for the system to succeed.

If we let Ej denote the event that the jth unit in Figure 11.2 is successful, then the series system reliability is expressed by [5]

R P E E E ES m

= ( ... )1 2 3 (11.13)

whereRS is the series system reliability. P(E1E2E3…Em) is probability of occurrence of events E1, E2, E3,…, and Em

For independent units, Equation (11.13) becomes

R P E P E P E P ES m

= ( ) ( ) ( )... ( )1 2 3 (11.14)

where P(Ej) is probability of occurrence of event Ej for j = 1, 2, 3,…, m.If we let Rj = P(Ej) for j = 1, 2, 3,…, m, Equation (11.14) becomes

R R R R RS m

=1 2 3

... (11.15)

where Rj is the unit j reliability for j = 1, 2, 3,…, m.For constant failure rate, lj, of unit j, using Equation (11.1), the reliability of

the unit j is given by

R t e

e

j

dt

t

tj

j

( ) =

=

− ∫

−

0 λ

λ

(11.16)

where Rj (t) is reliability of unit j at time t.

321 m

FIGURE 11.2Block diagram of a series system containing m units.

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Thus, by inserting Equation (11.16) into Equation (11.15), we obtain

R t eS

tjm

j( ) =− =Σ 1λ

(11.17)

where RS(t) is series system reliability at time t.By substituting Equation (11.17) into Equations (11.4) and (11.11), we get the

following equations for the series system mean time to failure and hazard rate, respectively:

MTTF e dtS

t

jm

j

jm

j=

=∑

−∞

=

=∫ Σ 1

0

1

1

λ

λ

(11.18)

and

λ λ

λ

λ

λ

St

jj

m t

e

ej

j

m

jj

m

= −∑

−

∑

=

− =

−

=

=∑1

1

1

1

jjj

m

=∑

1

(11.19)

whereMTTFS is series system mean time to failure.lS is series system hazard or failure rate.

Example 11.4Assume that an engineering system is composed of three independent subsystems in series. The failure rates of subsystems 1, 2, and 3 are 0.005 failure/hour, 0.004 failure/hour, and 0.003 failure/hour, respectively. Calculate the following:

engineering system reliability during a 40-hour mission;•engineering system mean time to failure; and•engineering system hazard rate.•

By substituting the specified data values into Equations (11.17), (11.18), and (11.19), we get

R e

MTT

S ( )

. ,

( . . . )( )40

0 6188

0 005 0 004 0 003 40==

− + +

FFS =+ +

=

10 005 0 004 0 003

83 33( . . . )

. hours,

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and

λS = + +=

( . . . )

.

0 005 0 004 0 003

0 012failures/hour

Thus, the engineering system reliability, mean time to failure, and hazard rate are 0.6188, 83.33 hours, and 0.012 failures/hour, respectively.

11.5.2 Parallel Network

In this case, all units are active and at least one of these units must function nor-mally for the system success. The block diagram of a parallel system contain-ing m units is shown in Figure 11.3. Each block in the figure denotes a unit.

If we let Ej denote the event that the jth unit in Figure 11.3 is unsuccessful, then the parallel system failure probability is given by [5]

F P E E E

p m= ( ... )

1 2 (11.20)

whereFp is parallel system failure probability.P E E E

m( ... )

1 2 is probability of occurrence of failure events E E E

m1 2, ,..., .

For independent units, Equation (11.20) becomes

F P E P E P E

p m= ( ) ( )... ( )

1 2 (11.21)

where P Ej

( ) is probability of occurrence of failure event Ej; j = 1, 2,…, m.

If we let Fj = P E j( ) for j = 1, 2,…, m in Equation (11.21) and then subtract the resulting equation from unity, we get the following expression for the paral-lel system reliability:

R F F F

p m= −1

1 2...

(11.22)

1

m

2

FIGURE 11.3Block diagram of a parallel system containing m units.

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whereRp is parallel system reliability.Fj is failure probability of unit j for j = 1, 2,…, m.

For constant failure rate, lj, of unit j, by subtracting Equation (11.16) from unity and then substituting it into Equation (11.22), we get

R t ep

t

j

mj( ) ( )= − −

−

=∏1 1

1

λ

(11.23)

where Rp (t) is parallel system or network reliability at time t.For identical units, by substituting Equation (11.23) into Equation (11.4), we

get the following expression for the parallel system or network mean time to failure:

MTTF e dt

j

pt m

j

m

= − −

=

−∞

=

∫

∑

[ ( ) ]1 1

1 1

0

1

λ

λ

(11.24)

whereMTTFp is mean time to failure of the parallel system with identical units. l is unit failure rate.

Example 11.5An engineering system is composed of three independent, active, and identical units; at least one of the units must operate normally for system success. The unit failure rate is 0.0002 failure/hour. Calculate

engineering system reliability for a 100-hour mission; and•engineering system mean time to failure.•

By substituting the given data values into Equations (11.23) and (11.24), we get

R e ep ( ) [ ][( . )( ) ( . )(100 1 1 10 0002 100 0 0002= − − −− − 1100 0 0002 1001

0 9406

) ( . )( )][ ]

.

−

=

−e

and

MTTFp = + +

=

10 0002

112

13

9 166 7

( . )

, . hours

Thus, the engineering system reliability and mean time to failure are 0.9406 and 9,166.7 hours, respectively.

K10869_Book.indb 153 8/26/09 2:12:12 PM


11.5.3 K-out-of-m Network

In this case, all m units are active and at least K units out of these m units must work normally for the system success. The parallel and series networks are the special cases of this network for K = 1 and K = m, respectively.

Using the binomial distribution for independent and identical units, the K-out-of m network or system reliability is expressed by [5]

R R RK

m j

mj m j

j K

m

=

− −

=∑ ( )1

(11.25)

where

j

m mm j j

=−

!( )! !

(11.26)

R is unit reliability.RK/m is K-out-of-m network or system reliability.

For constant failure rate, l, of each unit, by substituting Equation (11.3) into Equation (11.25), we get

R t e eK

m j

mj t t m j

j K

m

( ) ( )=

−− − −

=∑ λ λ1

(11.27)

where RK/m(t) is K-out-of-m network or system reliability at time t.Using Equation (11.27) in Equation (11.4) yields

MTTF e eK

m j

mj t t m j

j K

m

=

−

− − −

=∑ λ λ( )1

=

∞

=

∫

∑0

1 1

dt

jj K

m

λ

(11.28)

where MTTFK/m is K-out-of-m network or system mean time to failure.

Example 11.6Assume that an engineering system is composed of four independent and identical units in parallel. At least two units must operate normally for the system’s success. Calculate the engineering system mean time to failure if the failure rate of each unit is 0.0008 failure/hour.

K10869_Book.indb 154 8/26/09 2:12:38 PM



MTTFj

j2

42

41

0 00081

10 0008

12

13

14

=

= + +

=∑( . )

( . )

= 1354 16. hours

Thus, the engineering system mean time to failure is 1354.16 hours.

11.5.4 Standby System

This is another type of redundancy used to improve system reliability. In this case, the system has a total of (m + 1) units, but only one unit operates. The remaining m units are kept in their standby mode. As soon as the oper-ating unit fails, the switching mechanism detects the failure and turns on one of the m standby units.

The system fails when all the standby units fail. The system reliability for independent and identical units, time-dependent unit failure rate, and per-fect switching mechanism and standby units is given by [5]

R t t dt eSS

tj

t dtt

( ) ( ) ( )=

∫ − ∫λ λ

0

0 =

∑j

m

j0

/ !

(11.29)

whereRSS (t) is standby system reliability at time t.m is total number of standby units.l(t) is unit time-dependent failure rate or hazard rate.

For constant unit failure rate (i.e., l(t) = l), Equation (11.29) becomes

R t t e jSS

j t

j

m

( ) ( ) != −

=∑ λ λ /

0

(11.30)

By inserting Equation (11.30) into Equation (11.4), we get

MTTF t e j dt

m

SSj t

j

m

=

= +

−

=

∞

∑∫ ( ) !λ

λ

λ /00

1

(11.31)

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Example 11.7A standby system has three independent and identical units: one operating and two on standby. The failure rate of each unit is 0.0002 failure/hour. Calculate the standby system mean time to failure if the standby units remain as good as new in their standby mode and the switching mechanism is perfect.


MTTFSS = +

=

( )( . )

,

2 10 0002

15 000 hours

Thus, the standby system mean time to failure is 15,000 hours.

11.6 Reliability and Maintainability Relationship

In order to have a clear understanding of the relationship between reliability and maintainability, some of the important aspects of both reliability and maintainability are discussed next.

11.6.1 Reliability

This is a design characteristic that results in durability of the system or item to perform its specified mission subject to stated conditions and time period. It is accomplished through actions such as choosing optimum engi-neering principles, satisfactory component sizing, controlling processes, and testing. The following are some specific general principles of reliabil-ity [5,17]:

Design to minimize the occurrence of failures.•Provide fail-safe designs.•Design for simplicity.•Provide redundancy when required.•Use fewer numbers of parts to perform multiple functions.•Minimize stress on parts.•Provide for simple periodic adjustment of parts subject to wear.•Maximize the use of standard parts.•Use parts with proven reliability.•Provide satisfactory safety factors between strength and peak •stress values.

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11.6.2 Maintainability

This is a built-in design and installation characteristic. It provides the end product an inherent ability to be maintained, thus ultimately leading to improved mission availability and reduction in maintenance cost, required tools and equipment, and required man-hours and skill levels. Some of the specific general principles of maintainability include [5,17]:

Reduce life cycle maintenance costs.•Reduce the amount, frequency, and complexity of required mainte-•nance tasks.Reduce mean time to repair.•Establish the extent of preventive maintenance to be performed.•Reduce or eliminate altogether the need for maintenance.•Reduce the amount of supply supports required.•Consider benefits of modular replacement versus part repair or •throwaway.Provide for maximum interchangeability.•

11.7 Maintainability Measures

Various maintainability measures are used during the design phase to pro-duce effective products with respect to maintainability. Some of these mea-sures are mean time to repair (MTTR), the probability of completing repair in given time interval (i.e., the maintainability function); mean preventive maintenance time; and maximum corrective maintenance time. All these measures are presented next [5,7,17,18].

11.7.1 Mean Time to Repair (MTTR)

This is probably the most widely used maintainability measure or parameter in maintainability analysis. It is sometimes called mean corrective mainte-nance time. The system mean time to repair is defined by [5]

MTTR ti i

i

m

ii

m

=

= =

∑ ∑λ λ1 1

(11.32)

whereMTTR is system mean time to repair.m is number of units.li is constant failure rate of unit i for i = 1, 2, 3,…, m.ti is time required to repair unit i for i = 1, 2, 3,…, m.

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Example 11.8Assume that an engineering system is composed of three nonidentical subsys-tems—1, 2, and 3—with constant failure rates l1 = 0.0006 failure/hour, l2 = 0.0005 failure/hour, and l3 = 0.0004 failure/hour, respectively. Corrective maintenance times of subsystems 1, 2, and 3 are 4, 3, and 2 hours, respectively. Calculate the engineering system mean time to repair.


MTTR = + +( . )( ) ( . ) ( ) ( . ) ( )( .

0 0006 4 0 0005 3 0 0004 20 00006 0 0005 0 0004

3 133+ +

=. . )

. hours

Thus, the engineering system mean time to repair is 3.133 hours.

11.7.2 Maintainability Function

This is concerned with determining the probability of completing repair in a specified time interval. For a known repair time distribution, the maintain-ability function can be obtained by using the following equation [5,18]:

m t f t dtr

t

( ) ( )= ∫0

(11.33)

wherem(t) is maintainability function (i.e., the probability that repair will be

accomplished in time t when it starts at time t = 0).t is time.fr(t) is probability density function of the repair times.

Example 11.9Assume that the repair times of a system are defined by the following probability density function (i.e., the repair times are exponentially distributed):

f t

MTTRer

tMTTR( )= −

1 1 (11.34)

where fr (t) is probability density function of the system repair times. t is time. MTTR is system mean time to repair.

Obtain an expression for the maintainability function and calculate the probability that a repair will be accomplished in 3 hours if the system mean time to repair is 4 hours.

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m tMTTR

e dt

e

MTTR

MT

tt

( ) =

= −

−

−

∫ 1

1

1

1

0

TTRt

(11.35)

Using the given data values in Equation (11.35) yields

m e( )

.

( )3 1

0 5276

14

3= −=

−( )

Thus, the expression for the maintainability function is given by Equation (11.35) and the probability that the system repair will be accomplished in 3 hours is 0.5276.

11.7.3 Mean Preventive Maintenance Time

This is a quite useful maintainability measure expressed by [5,18]

T t f fmp pi pi

i

K

pii

K

=

= =

∑ ∑1 1

(11.36)

whereTmp is mean preventive maintenance time.K is number of preventive maintenance tasks.tpi is elapsed time for preventive maintenance task i for i = 1, 2, 3,…, K.fpi is frequency of preventive maintenance task i for i = 1, 2, 3,…, K.

During the computation of Tmp, note that if the frequencies, fpi, are specified in maintenance tasks per hour, then the values of tpi must also be expressed in hours.

11.7.4 Maximum Corrective Maintenance Time

This maintainability measure is concerned with estimating the time to com-plete a specified percentage of all potential repair actions. Usually, the speci-fied percentiles are the 90th and 95th. Because the estimation of maximum corrective maintenance time depends on the probability distribution describ-ing the times to repair, equations for estimating maximum corrective main-tenance time for three probability distributions are presented next [5,18].

11.7.4.1 Exponential

In this case, the maximum corrective maintenance time is expressed by

MT MTTRCm

=α ( ) (11.37)

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whereMTCm is maximum corrective maintenance time.MTTR is mean time to repair.a is a constant whose values are 2.312 and 3 for the 90th and 95th percen-

tiles, respectively.

11.7.4.2 Normal

In this case, the maximum corrective maintenance time is defined by

MT MTTRCm n

= +θ σ (11.38)

whereq is a constant and its values are 1.28 and 1.65 for the 90th and 95th percen-

tiles, respectively.sn is standard deviation of the repair times.

11.7.4.3 Lognormal

In this case, the maximum corrective maintenance time is expressed by

MT anti TCm a

= +log ( )θ σl

(11.39)

whereTa is mean of the logarithms of repair times.σ

l is standard deviation of the logarithms of the repair times.

Additional information on the maximum corrective maintenance time is available in Dhillon [5,8].

11.8 System Availability and Unavailability

Availability and unavailability of a system are given by [5,18]

AV t e

SS S

S StS S( )

( )[ ]( )=

++ − +1

λ µµ λ λ µ (11.40)

and

UAV t e

SS

S S

tS S( ) [ ]( )=+

− − +λλ µ

λ µ1 (11.41)

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whereAVS (t) is system availability at time t.t is time.lS is system constant failure rate.mS is system constant repair rate.UAVS (t) is system unavailability at time t.

As time t becomes very large, Equations (11.40) and (11.41) reduce to

AV AV t

S t SS

S S

= =+→∞

lim

( )µ

λ µ (11.42)

and

UAV UAV t

S t SS

S S

= =+→∞

lim

( )λ

λ µ (11.43)

whereAVS is system steady-state availability.UAVS is system steady-state unavailability.

Because

λ

SS

MTTF= 1

and

µ

SS

MTTR= 1

Equations (11.42) and (11.43) become

AV

MTTFMTTF MTTR

System uptimeSystem uptiS

S

S S

=+

=mme System downtime+

(11.44)

and

UAV

MTTRMTTF MTTR

System downtimeSystem uS

S

S S

=+

=pptime System downtime+

(11.45)

whereMTTFS is system mean time to failure.MTTRS is system mean time to repair.

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Example 11.10An engineering system mean time to failure and mean time to repair are 400 hours and 20 hours, respectively. Calculate the system steady-state unavailability.


UAVS =

+=20

400 200 0476.

Thus, the engineering system unavailability is 0.0476.

11.9 Reliability and Maintainability Tools

Many methods are used to perform various types of reliability and main-tainability analyses. Three of these methods that can be used to perform both reliability and maintainability analyses are as follows:

failure modes and effect analysis (FMEA);•fault tree analysis; and•cause and effect diagram.•

Each of these methods is described next.

11.9.1 Failure Modes and Effect Analysis (FMEA)

This is an important tool to evaluate engineering design at the initial stage from the reliability and maintainability aspects. FMEA was developed in the early 1950s for evaluating the design of flight control systems [19].

It helps to identify the need for and effects of design change and demands listing of potential failure modes of all system or equipment components on paper and their effects on the listed subsystems. The main steps in per-forming FMEA are shown in Figure 11.4. FMEA is called failure modes, effects, and criticality analysis (FMECA) when criticalities are assigned to failure mode effects. Additional information on FMEA is available in Dhillon [5].

11.9.2 Fault Tree Analysis

This is one of the most widely used methods for performing system reli-ability analysis; it arranges fault events in a tree-shaped diagram (thus, the name). The method is well suited to determine the combined effects of mul-tiple failures. It was originally developed to evaluate the reliability of the Minuteman launch control system at Bell Telephone Laboratories in the early 1960s [5,20].

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The fault tree analysis starts by identifying an undesirable event—called the “top event”—associated with a system under consideration. The fault events that can cause the occurrence of the top event are generated and connected by logic gates known as OR, AND, etc. The construction of a fault tree of a system proceeds by generation of fault events (by asking, “How can this event occur?”) successively until the fault events need not be developed further. These events are called elementary or primary events.

Overall, a fault tree may simply be described as the logic structure relating the primary fault events to the top event. Additional information on fault tree analysis is available in Dhillon [5] and Dhillon and Singh [21].

11.9.3 Cause and Effect Diagram

This is a quite useful approach for determining the root cause of a given problem and generating relevant ideas. Other names used for this approach are Ishikawa diagram, after its Japanese originator K. Ishikawa, and “fish bone” diagram because of its resemblance to the skeleton of a fish (as shown in Figure 11.5).

It can be seen from this figure that the right side (i.e., the fish head or the box) represents the effect (the problem or goal) and the dotted box on the left side contains “fish bones” that can be any set of factors considered to be important causes.

Step 1

Step 2

Step 3

Step 4

Step 5

Step 6

Step 7

Define all system boundaries and detailed requirements

Identify and describe each component and list all its associated failure modes

Assign failure probability/rate to each failuremode

List effects of each failure mode on subsystem/system/plant

Enter necessary remarks for each failuremode

Review each failure mode considered criticaland take appropriate actions

List all components/subsystems in the systemunder consideration

FIGURE 11.4Steps for performing FMEA.

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The following basic steps are used to develop a cause and effect diagram:

Develop problem statement.•Brainstorm to identify all possible causes.•Develop important cause classifications by stratifying into natural •groupings and process steps.Develop the diagram.•Refine all cause classifications by asking questions such as “What •causes this?” and “Why does this condition exist?”

Additional information on the cause and effect diagram is available in Dhillon [5,18].

Problems

1. Discuss reliability and maintainability history. 2. Define the following terms:

availability;•reliability;•maintainability; and•useful life.•

3. Describe the bathtub hazard rate curve. 4. Write three different general formulas for obtaining mean time to

failure. 5. Obtain an expression for a parallel system hazard rate by using

Equation (11.23). 6. List at least 10 general principles of reliability.

Effect (problem or goal)

“Fish head”

Cause (n-1) Cause 3 Cause 1

Cause n Cause 4 Cause 2

FIGURE 11.5Cause and effect diagram layout.

K10869_Book.indb 164 8/26/09 2:14:58 PM


7. Describe two methods that can be used to perform reliability and maintainability analyses.

8. Prove that the sum of Equations (11.40) and (11.41) is equal to unity.

9. Assume that an engineering system is composed of four noniden-tical subsystems—1, 2, 3, and 4—with constant failure rates l1 = 0.0001 failure/hour, l2 = 0.0002 failure/hour, l3 = 0.0003 failure/hour, and l4 = 0.0004 failure/hour, respectively. Calculate the engineering system mean time to repair if the corrective mainte-nance times of subsystems 1, 2, 3, and 4 are 2, 4, 6, and 8 hours, respectively.

10. Assume that an engineering system is composed of five indepen-dent and identical units in parallel. At least three units must operate normally for the system success. Calculate the engineering system mean time to failure if the constant failure rate of each unit is 0.004 failure/hour.

References

1. Lyman, W. J. 1933. Fundamental consideration in preparing a master system plan. Electrical World 101:778–792.

2. Smith, S. A. 1934. Service reliability measured by probabilities of outage. Electrical World 103:371–374.

3. Dhillon, B. S. 1983. Power system reliability, safety, and management. Ann Arbor, MI: Ann Arbor Science Publishers.

4. Coppola, A. 1984. Reliability engineering of electronic equipment: A historical perspective. IEEE Transactions on Reliability 33:29–35.

5. Dhillon, B. S. 1999. Design reliability: Fundamentals and applications. Boca Raton, FL: CRC Press.

6. Dhillon, B. S. 2007. Applied reliability and quality: Fundamentals, methods, and pro-cedures. London: Springer–Verlag.

7. AMCP 706-133. 1976. Engineering design handbook: Maintainability engineering theory and practice. Washington, D.C.: Department of Defense.

8. Moss, M. A. 1985. Minimal maintenance expense. New York: Marcel Dekker, Inc. 9. Retterer, B. L., and Kowalski, R. A. 1984. Maintainability: A historical perspec-

tive. IEEE Transactions on Reliability 33:56–61. 10. Shooman, M. L. 1968. Probabilistic reliability: An engineering approach. New York:

McGraw–Hill Book Company. 11. Dhillon, B. S. 1993. Reliability and quality control: Bibliography on general and spe-

cialized areas. Gloucester, Ontario, Canada: Beta Publishers, Inc. 12. Dhillon, B. S. 1993. Reliability engineering applications: Bibliography on important

application areas. Gloucester, Ontario, Canada: Beta Publishers, Inc. 13. MIL-STD-721. 1974. Definitions of effectiveness terms for reliability, maintain-

ability, human factors, and safety. Washington, D.C.: Department of Defense.

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14. Naresky, J. J. 1970. Reliability definitions. IEEE Transactions on Reliability 19:198–200.

15. Omdahl, T. P., ed. 1988. Reliability, availability, and maintainability (RAM) diction-ary. Milwaukee, WI: ASQC Quality Press.

16. Von Alven, W. H., ed. 1964. Reliability engineering. Englewood Cliffs, NJ: Prentice Hall, Inc.

17. AMCP-706-134. 1972. Maintainability guide for design. Prepared by the Department of the Army, Department of Defense, Washington, D.C.

18. Dhillon, B. S. 1999. Engineering maintainability: How to design for reliability and easy maintenance. Houston, TX: Gulf Publishing Company.

19. Countinho, J. S. 1964. Failure effect analysis. Transactions of the New York Academy of Sciences 26:564–584.

20. Haasl, D. F. 1965. Advanced concepts in fault tree analysis. Proceedings of the System Safety Symposium. Available from the University of Washington Library, Seattle, WA.

21. Dhillon, B. S., and Singh, C. 1981. Engineering reliability: New techniques and appli-cations. New York: John Wiley & Sons.

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167

Bibliography: Literature on Life Cycle Costing

Introduction

Over the years, a large number of publications on various aspects of life cycle costing have appeared in the form of journal articles, conference proceed-ings articles, books, etc. This bibliography presents an extensive list of such publications. The period covered by the listing is from 1988 to 2008. The main objective of this listing is to provide readers with sources for obtaining additional information on life cycle costing.

Publications

Abraham, D. M. 2003. Life cycle cost integration for the rehabilitation of wastewater infrastructure. Proceedings of the Construction Research Congress 627–635.

Adler, D., Willman, T., and Lilly, E. 1997. Figuring life-cycle costs in the real world. Chemical Processing 60 (8): 29–32.

Adler, D. J., Herkamp, J. A., Wiesler, J. R., and Williams, S. B. 1995. Life cycle cost and benefits of process automation in bulk pharmaceuticals. ISA Transactions 34 (2): 133–139.

Ahmed, N. U. 1995. A design and implementation model for life cycle cost manage-ment system. Information and Management 28, (4): 261–269.

Akselsson, H., and Burstrom, B. 1994. Life cycle cost procurement of Swedish State Railways’ high-speed train X2000. Proceedings of the Institution of Mechanical Engineers, Part F: Journal of Rail and Rapid Transit 208 (1): 51–59.

Aktacir, M. et al. 2006. Life-cycle cost analysis for constant-air-volume and variable-air-volume air-conditioning systems. Applied Energy 83 (6): 606–627.

Alfredsson, K. 2001. Life cycle cost in focus. Water and Wastewater International 16 (2): 25.Ali Khan Malik, M., and Kolodchak, P. 1990. Cost-reliability relationship in life cycles.

Proceedings of the International Industrial Engineering Conference 581–586.Allen, E. C. et al. 2006. Mission-based simulation software development for optimiz-

ing air vehicle life cycle costs. Proceedings of the AIAA Modeling and Simulation Technologies Conference 1021–1031.

Anon. 1993. Effect of life cycle costs versus initial costs. NASA Reference Publication 1310:146.

———. 1994. Life cycle costing. Steel Times 222 (1): 21–22.———. 1995. Improving gas turbine availability and life cycle costs. International

Power Generation 18 (5): 38–39, 42.

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168 Bibliography: Literature on Life Cycle Costing

———. 1995. Life cycle costing: Report on the introduction of life cycle costing tech-niques to the selection of maintenance coating systems for offshore fabric. Offshore Engineer 7:30.

———. 1995. Reader responds to life cycle costing article. Chemical Processing 58 (8): 12.———. 1995. Thermal insulation environmental impacts and life-cycle costs. Construction

Specifier 48 (6): 64–69.———. 1996. Do life-cycle costs validate the standard-plant concept? Power 140 (8): 126.———. 1996. Life-cycle costing reveals masonry’s long-term value. Aberdeen’s Magazine

of Masonry Construction 9 (12): 555.———. 1996. Up time: New low-maintenance components from Eaton help reduce

operating and life cycle costs. Diesel Equipment Superintendent 74 (4): 58.———. 1997. GTX100 promises high reliability and low life cycle costs. Modern Power

Systems 17 (7): 23, 25.———. 1997. Life cycle cost implications of roofing decisions. Interface 15 (2): 7.———. 1997. Life cycle costing proves concrete’s economy. Better Roads January: 21–24.———. 1997. Life-cycle costing provides economy. Better Roads 67 (1): 21.———. 1997. Life cycle costs. Aerospace Engineering 17 (10): 29.———. 1998. Bridge plans receive life-cycle costs. ENR (Engineering News-Record) 240

(20): 19.———. 1998. It’s time to calibrate financial models with real life-cycle costs. Power

142 (4): 4.———. 1998. Life cycle cost analysis for pumping systems. World Pumps 383:28–32.———. 1999. Procedures for welding titanium piping helping U.S. Navy to reduce

ship life-cycle costs. Welding Journal 78 (4): 92.———. 1999. State DOTs update life-cycle cost analysis. Better Roads 69 (10): 25.———. 2000. Intelligent wells: Forecasting life-cycle costs. Hart’s E and P 73 (8): 125.———. 2000. ITT: Technology leadership and customer satisfaction driving life cycle

cost. World Pumps April: 18–21.———. 2000. Managing equipment life-cycle costs. Chemical Engineering 107 (2): 80.———. 2000. Pump users’ forum with a focus on life cycle costs. World Pumps 407:44.———. 2001. Life-cycle strategy for pumps improves cost structure. World Pumps

413:30–32.———. 2001. Roofing and life-cycle cost. Buildings 95 (5): 74.———. 2002. A local authority thinks hard about life-cycle costs. Highways 71 (5): 32.Arditi, D. A. et al. 1996. Life-cycle costing in municipal construction projects. Journal

of Infrastructure Systems 2 (1): 5–14.Arditi, D., and Messiha, H. M. 1999. Life cycle cost analysis (LCCA) in municipal

organizations. Journal of Infrastructure Systems 5 (1): 1–10.Arnold, B. D. et al. 2005. Life-cycle costing of air filtration. ASHRAE Journal 47 (11):

30–32.Arpke, A., and Hutzler, N. 2005. Operational life-cycle assessment and life-cycle

cost analysis for water use in multi-occupant buildings. Journal of Architectural Engineering 11 (3): 99–109.

Ashworth, A. 1989. Life-cycle costing: A practice tool? Cost Engineering 31 (3): 8–11.Asiedu, Y., and Gu, P. 1998. Product life cycle cost analysis: State of the art review.

International Journal of Production Research 36 (4): 883–908.Balda, D. M., and Gustafson, D. A. 1990. Cost estimation models for the reuse and

prototype software development life-cycles. ACM SIGSOFT Software Engineering Notes 15 (3): 42–50.

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Bibliography: Literature on Life Cycle Costing 169

Baliwangi, L. et al. 2006. Optimizing ship machinery maintenance scheduling through risk analysis and life cycle cost analysis. Proceedings of the 25th International Conference on Offshore Mechanics and Arctic Engineering 127–133.

Bang, K. L. et al. 1996. Development of guidelines based on life-cycle cost to replace level-of-service concept in capacity analysis. Transportation Research Record 1572:9–17.

Barrick, M. D. 1989. Productivity and life cycle cost issues in applications of embedded fiber optic sensors in smart skins. Proceedings of the SPIE Conference 171–179.

Barros, L. L. 1998. The optimization of repair decisions using life cycle cost param-eters. IMA Journal of Mathematics Applied in Business and Industry 9 (4): 403–413.

Battlebury, D. R. 1991. The practical application of life cycle costing to the design of power systems. Proceedings of the Third International Conference on Probabilistic Methods Applied to Electric Power Systems 6–8.

Bears, J., and Coathup, L. 1991. Evaluation of the life cycle cost for universal fiber access. Proceedings of the First International Workshop on Photonic Networks, Components and Applications 81–89.

Becker, E. A. et al. 1999. Life cycle cost of urban pavements. Concrete Engineering International 3 (3): 26–28.

Becker, S. 1998. Bringing advanced bogie technology to Europe will cut life-cycle costs. Railway Gazette International 154 (9): 599–600.

Bell, J. H. 1990. Parts recovery life cycle costs. Proceedings of the Test Engineering Conference 181–185.

Bell, P. I., and Trigger, J. P. 1998. Access network life-cycle costs. BT Technology Journal 16 (4): 165–174.

Bentz, E. J., Bentz, C. B., and O’Hora, T. D. 2001. Comparative assessment of low-level radioactive waste life-cycle disposal costs of U.S. commercial facilities. Proceedings of the 8th International Conference on Radioactive Waste Management and Environmental Remediation 751–757.

Bescherer, F. 2006. Towards the optimum cost of ownership of switched-mode power supplies: Early stage cost management with life-cycle costing. Proceedings of the IEEE 32nd Annual Conference on Industrial Electronics 2203–2207.

Bettigole, N. H. 1993. Bridge engineering and life cycle cost. Proceedings of the Structures Congress 1047–1052.

———. 1995. Bridge management and life cycle cost. Proceedings of the Structures Congress 668–669.

Bhaskaran, R., Palaniswamy, N., and Rengaswamy, N. S. 2006. Life-cycle cost analysis of a concrete road bridge across open sea. Materials Performance 45 (10): 51–55.

Birkenshaw, J. 2003. Life cycle costing of print on-demand digital printing of books and packaging materials. Proceedings of the International Conference on Digital Production Printing and Industrial Applications 12–13.

Blanchard, B. S. 1988. The measures of a system—performance, life-cycle cost, system effectiveness, or what? Proceedings of the IEEE National Aerospace and Electronics Conference 1434–1439.

Bodsberg, L., and Hokstad, P. 1995. A system approach to reliability and life-cycle cost of process safety systems. IEEE Transactions on Reliability 44 (2): 179–186.

Boehm, B. et al. 2004. A software product line life cycle cost estimation model. Proceedings of the International Symposium on Empirical Software Engineering 156–164.

Bohoris, G. A. 1993. Life-cycle costs and comparative statistical techniques for cen-sored reliability data. Journal of the Operational Research Society 44 (4): 355–360.

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Bonner, J. A. et al. 1989. Fossil power plant life cycle management as a least cost plan-ning approach available to utility and industrial plant operators. Proceedings of the American Power Conference 956–958.

Botelho, D. et al. 2000. Life-cycle-cost-based design criteria for Gulf of Mexico mini-mum structures. Proceedings of the Annual Offshore Technology Conference 87–94.

Boussabaine, H. A., and Kirkham, R. J. 2004. Whole life cycle costing: Risk and risk responses. Oxford, England: Blackwell Publishing.

Breidenbach, D. P. 1989. Life cycle cost analysis. Proceedings of the IEEE 1989 National Aerospace and Electronics Conference 1216–1220.

Breniere, X., and Tribolet, P. 2006. IR detectors life cycle cost and reliability optimiza-tion for tactical applications. Proceedings of the International Society for Optical Engineering Conference on Electro-Optical and Infrared Systems: Technology and Applications III 63950D1–63950D12.

Brentlinger, L. A., Hofmann, P. L., Peterson, R. W., and Dippold, D. G. 1988. Transportable storage casks: An analysis of life cycle dose and life cycle cost. Proceedings of the Summer Computer Simulation Conference 742–746.

Brooks, S. M. 1996. Life cycle costs estimates for conceptual ideas. Proceedings of the IEEE National Aerospace and Electronics Conference 541–546.

Brown, D. R., and Humphreys, K. K. 1988. Battery life-cycle cost analysis. Electric Vehicle Developments 7 (3): 81–82.

Bruhwiler, E., and Adey, B. 2005. Improving the consideration of life-cycle costs in bridge decision-making in Switzerland. Structure and Infrastructure Engineering 1 (2): 145–157.

Bruzzone, A. G., Briano, C., Massei, M., and Poggi, S. 2006. Simulation and optimi-zation as decision support system in relation to life cycle cost of new aircraft carriers. Proceedings of the Sixth IASTED International Conference on Modeling, Simulation, and Optimization 133–138.

Buncher, M., and Rosenberger, C. 2006. Understanding the true economics of using polymer modified asphalt through life cycle cost analysis. Paving the Way 8 (2): 1–20.

Burley, E., and Rigden, S. R. 1997. Use of life cycle costing in assessing alternative bridge design. Proceedings of the Institution of Civil Engineers 121 (1): 22–27.

Burns, D. J., and Formaniak, A. 1992. Reliability and life cycle cost evaluation for sys-tem design. Proceedings of the Safety and Reliability Conference 353–367.

Burrows, C. 2003. Designed for life—First cost or life. Proceedings of the International Conference on New Trains 19–26.

Burstrom, B., Ericsson, G., and Kjellsson, U. 1994. Verification of life-cycle cost and reliability for the Swedish high speed train X2000. Proceedings of the Annual Reliability and Maintainability Symposium 166–171.

Cain, J. P., Habash, N., and Gibson, J. A. 1994. Analysis of military systems using an interactive life cycle costing model. Proceedings of the IEEE National Aerospace and Electronics Conference 1218–1224.

Calvo, A. B., Danish, A. J., and Marcus, D. 2002. Web-LCCA: A life-cycle cost model for evaluation of COTS and custom display designs. Proceedings of SPIE Conference 70–80.

Cardullo, M. W. 1993.Total life-cycle cost analysis of conventional and alternative fueled vehicles. IEEE Aerospace and Electronic Systems Magazine 8 (11): 39–43.

———. 1995. Total life cycle cost model for electric power stations. Proceedings of the 30th Intersociety Energy Conversion Engineering Conference 409–414.

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Carnahan, J. V., and Marsh, C. 1998. Comparative life-cycle cost analysis of under-ground heat distribution systems. Journal of Transportation Engineering 124 (6): 594–605.

Carriere, M., Schoenau, G. J., and Besant, R. W. 1998. Revised procedure for duct design with minimum life-cycle cost. ASHRAE Transactions 104 (2): 62–67.

Carrubba, E. R. 1992. Integrating life-cycle cost and cost of ownership in the com-mercial sector. Proceedings of the Annual Reliability and Maintainability Symposium 101–108.

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Chafee, S. S. 1996. Fundamental requirements of life cycle costing: Projecting life cycle costs for electronic system modernization. Proceedings of the IEEE Technical Applications Conference 41–46.

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Chang, S. E., and Shinozuka, M. 1996. Life-cycle cost analysis with natural hazard risk. Journal of Infrastructure Systems 2 (3): 118–126.

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Christian, J. et al. 1998. Life cycle costs of the barrack block: Are we building better and smarter? Proceedings of the Annual Conference of the Canadian Society for Civil Engineering 129–138.

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Cranford, E. L., III et al. 2002. Reduced life cycle costs and improved analysis accu-racy utilizing Westem’s integrated modeling methods. ASME Pressure Vessels and Piping Division Publication (PVP) 443 (1): 9–15.

Crawley, M. F., and Bell, J. M. 1993. Application of life-cycle cost analysis to pneu-matic conveying systems. Powder Handling & Processing 5 (3): 213–218.

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———. 1993. FALCCM-H: Functional avionics life cycle cost model for hardware. Proceedings of the IEEE National Aerospace and Electronics Conference 950–953.

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De Vasconcellos, N. M., and Yoshimura, M. 1999. Life cycle cost model for acqui-sition of automated systems. International Journal of Production Research 37 (9): 2059–2076.

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Dhillon, B. S. 1989. Life cycle cost: Techniques, models, and applications. New York: Gordon and Breach Science Publishers.

Dieffenbach, J. R., and Mascarin, A. E. 1993. Body-in-white material systems: A life-cycle cost comparison. JOM 45 (6): 16–19.

Dietrich, J. M. 2004. Life cycle process management for environmentally sound and cost effective semiconductor manufacturing. Proceedings of the IEEE International Symposium on Electronics and the Environment 168–172.

Di Martino, P., Rosi, R., and Zanetta, L. 1996. Life cycle costs comparison and sensitiv-ity analysis for multimedia networks. Proceedings of the European Conference on Networks and Optical Communications 306–310.

Di Stefano, P. 2006. Tolerances analysis and cost evaluation for product life cycle. International Journal of Production Research 44 (10): 1943–1961.

Doswell, B. E. 1988. Who is doing what about life cycle costing. Proceedings of the Conference on Military Computers, Graphics and Software 333–344.

Dowdell, D. C. et al. 2000. An integrated life cycle assessment and cost analysis of the implications of implementing the proposed waste from electrical and electronic equipment (WEEE) directive. Proceedings of the IEEE International Symposium on Electronics and the Environment 1–10.

Dowlatshahi, S. 1997. Elements of time-based competition and life cycle costing in concurrent engineering environments. Proceedings of the Annual Meeting of the Decision Sciences Institute 1091–1092.

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197

Index

A

Aircraft airframe maintenance cost drivers, 109–110

Aircraft cost drivers, 108–110Aircraft life cycle cost, 105–107Aircraft turbine engine life cycle cost,

108American National Standards

Institute, 6American Public Power Association, 7American Society for Quality Control,

conference proceedings, 4American Society of Civil

Engineers, 6–7American Society of Heating,

Refrigeration and Air Conditioning Engineers, 7

American Society of Mechanical Engineers, 7

Analysis costas software life cycle cost, 97software life cycle cost element, 98

Analytic models, software cost estimation, 99

Annual American Society for Quality Control Conference, proceedings, 4

Annual Canadian Society for Civil Engineering Conference, proceedings, 4

Annual Conference of the Urban and Regional Information Systems Association, proceedings, 4

Annual Offshore Technology Conference, proceedings, 4

Annual Petroleum and Chemical Industry Conference, proceedings, 4

Annual Reliability and Maintainability Symposium, proceedings, 4

Annual Reliability Engineering Conference for the Electric Power Industry, proceedings, 4

ANSI. See American National Standards Institute

Appliance life cycle costing, 122–123Application areas, 28–29Areas for evaluation, 31–32ASCE. See American Society of Civil

EngineersASHRAE. See American Society of

Heating, Refrigeration and Air Conditioning Engineers

ASME. See American Society of Mechanical Engineers

ASQC. See American Society for Quality Control

Asset condition, experience, 78Automobile life cycle cost, 113–114

B

Bathtub hazard rate curve, 146–147Better Roads, 3Blanchard, B.S., 5Boussabaine, A., 5“Bridge Life Cycle Cost Analysis,” 5Bridge life cycle costs, 119–120Brown, R.J., 5Building energy cost estimation,

120–122formula I, 120–121formula II, 121formula III, 121–122formula IV, 122formula V, 122

Building life cycle cost, 117–118Bull, J.W., 5Bus life cycle cost estimation

model, 114

C

Canadian Society for Civil Engineering, conference proceedings, 4

Car life cycle cost, 113–114

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198 Index

Cargo ship life cycle cost, 110–111Certification cost, software life cycle

cost element, 98Chemical Engineering, 4Circuit-breaker life cycle cost

estimation model, 126City bus life cycle cost estimation

model, 114Civil engineering structures, energy

systems life cycle costing, 117–128

appliance life cycle costing, 122–123

bridge life cycle costs, 119–120building energy cost estimation,

120–122formula I, 120–121formula II, 121formula III, 121–122formula IV, 122formula V, 122

building life cycle cost, 117–118circuit-breaker life cycle cost

estimation model, 126energy system life cycle cost

estimation model, 123motor life cycle cost estimation

model, 124–125pump life cycle cost estimation

model, 125–126steel structure life cycle cost,

118–119waste treatment facilities life cycle

costs, 119–120Code and checkout cost

as software life cycle cost, 97software life cycle cost element, 98

Coded instructions, software life cycle cost element, 98

Combat aircraft hydraulic, fuel systems cost drivers, 110

Company policy, maintenance costing, 78

Compiling programs, software life cycle cost element, 98

Composite models, software cost estimation, 98–99

Compound amount, 3Compound interest, 12–14

Computer system life cycle costing, 91–104

analysis costas software life cycle cost, 97software life cycle cost

element, 98certification cost, software life cycle

cost element, 98code and checkout cost

as software life cycle cost, 97software life cycle cost

element, 98coded instructions, software life

cycle cost element, 98compiling programs, software life

cycle cost element, 98computer system life cycle costing,

software life cycle cost element, 98

data structure cost, software life cycle cost element, 98

design costas software life cycle cost, 97software life cycle cost

element, 98design requirements and

specifications, software life cycle cost element, 98

desk checks, software life cycle cost element, 98

documentation costas software life cycle cost, 97software life cycle cost

element, 98documentation revisions, software

life cycle cost element, 98environments, software life cycle

cost element, 98flow charts, software life cycle cost

element, 98input and output parameters,

software life cycle cost element, 98

installation costas software life cycle cost, 97software life cycle cost

element, 98interface requirements, software life

cycle cost element, 98

K10869_Book.indb 198 8/26/09 2:15:03 PM

Index 199

listing, software life cycle cost element, 98

maintenance costs, 93–95maintenance manual, software life

cycle cost element, 98models, 91–93modifications, software life cycle

cost element, 98operating and support cost


program requirements, software life cycle cost element, 98

program test, software life cycle cost element, 98

software cost estimationanalytic models, 99composite models, 98–99linear models, 99methods, 97–103models, 100–103multiplicative models, 99–100tabular models, 98

software costing, 95–96software life cycle cost

elements, 97influencing factors, 96–97

software life cycle cost element, 98subelements, 98

system integration, software life cycle cost element, 98

system requirements, software life cycle cost element, 98

test and integration costas software life cycle cost, 97software life cycle cost

element, 98test procedures, software life cycle

cost element, 98test revisions, software life cycle

cost element, 98user manual, software life cycle cost

element, 98validation cost, software life cycle

cost element, 98verification cost, software life cycle

cost element, 98Concrete Engineering International, 4Conference proceedings, 4

Contracts, incorporating into planning process for, 30–31

Cooling tower cost estimation model, 133

Corrective maintenance labor cost estimation, 80–81

Cost estimating relationship, 3Cost estimation methods, 55–58

method I, 55–56method II, 56–57method III, 57method IV, 57–58method V, 58

Cost models, 3, 43–61cost estimation methods, 55–58

method I, 55–56method II, 56–57method III, 57method IV, 57–58method V, 58

general life cycle cost models, 44–50model I, 44–45model II, 45–46model III, 46model IV, 47model V, 47–49model VI, 49–50

specific life cycle cost models, 50–55model I, 50–51model II, 51–52model III, 52–53model IV, 53–54model V, 55

types of life cycle cost models, inputs, 43–44

D

Data collection, maintenance cost, 85Data information sources, 6Data source, 32–33Data structure cost, software life cycle

cost element, 98Declining-balance method, 22–24Defense Management Journal, 4Defense Technical Information

Center, 6Dell’isola, A.J., 5Depreciation, defined, 20

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200 Index

Depreciation methods, 20–23Design cost


Design requirements, specifications, software life cycle cost element, 98

Desk checks, software life cycle cost element, 98

Development cost estimation model, 131–132

Dhillon, B.S., 5Documentation cost


Documentation revisions, software life cycle cost element, 98

Downtime, 2

E

Earles, M.E., 5Effective annual interest rate, 14–15“Energy Price Indices and Discount

Factors for Life Cycle Cost Analysis,” 5

Energy system life cycle cost estimation model, 123

Engineering reliability, maintainability, 145–166

bathtub hazard rate curve, 146–147common reliability networks,

150–156general reliability, 147–149hazard rate formulas, 147–149maintainability measures, 157–160maintainability tools, 162–164mean time to failure, 147–149reliability, 156–157reliability tools, 162–164system availability, unavailability,

160–162Environments, software life cycle cost

element, 98Equipment maintenance cost, 79–80Equipment selection, from competing

manufacturers, 34–40Equipment specification, maintenance

costing, 78

Estimation costs, reliability-related tasks, models, 64

model I, 64–66model II, 65model III, 65model IV, 65model V, 66

European Transactions on Electric Power, 4

F

Fabrycky, W.J., 5Failure, defined, 3Failure rate, 3Flow charts, software life cycle cost

element, 98Frangopol, D.M., 5

G

General life cycle cost models, 44–50model I, 44–45model II, 45–46model III, 46model IV, 47model V, 47–49model VI, 49–50

GIDEP. See Government Industry Data Exchange Program

Government Industry Data Exchange Program, 6

H

Hawk, H., 5Hazard rate formulas, 147–149Helicopter maintenance cost drivers, 09Hunkeler, D., 5

I

IEEE Aerospace and Electronic Systems Magazine, 4

IEEE Annual Conference on Industrial Electronics, proceedings, 4

IEEE Annual Pulp and Paper Industry Technical Conference, proceedings, 4

IEEE Transactions on Power Delivery, 4

K10869_Book.indb 200 8/26/09 2:15:03 PM

Index 201

IEEE Transactions on Reliability, 3Industrial sector quality cost

classifications distribution, 66–68

appraisal cost, 67external failure cost, 67–68internal failure cost, 67prevention cost, 67

Information and Management, 3Information required, 27–28Input/output parameters, software life

cycle cost element, 98Installation cost


Interestcompound, 12–14simple, 11–12

Interface requirements, software life cycle cost element, 98

International Journal of Production Economics, 4

International Journal of Production Research, 3

International Journal of Quality and Reliability Management, 4

International Power Generation, 3Investment cost elements,

maintainability, 85ISSAT International Conference on

Reliability and Quality in Design, proceedings, 4

J

Journal of Infrastructure Systems, 3Journal of Quality in Maintenance

Engineering, 3Journal of Transportation Engineering, 4

K

Kirk, S.J., 5Kirkham, R., 5

L

Lichtenvort, K., 5Life cycle cost, 2

“Life Cycle Cost Analysis and Design of Civil Infrastructure System,” 5

“Life Cycle Cost Analysis and Design of Civil Infrastructure System,” Structural Engineering Institute of the American Society of Civil Engineers, 5

“Life Cycle Cost Analysis of Pavements,” 5

“Life Cycle Cost in Navy Acquisitions,” 4

“Life Cycle Cost Model for Defense Material Systems Data Collection Workbook,” 4

Life cycle costing, contracts, incorporating into planning process for, 30–31

“Life Cycle Costing Manual: For the Federal Energy Management Program,” 5

Life cycle costing-related response, requirements analysis, 31

Linear models, software cost estimation, 99

Listing, software life cycle cost element, 98

M

Machining life cycle cost analyses, 35–40

Maintainability, 2investment cost elements, 85measures, 157–160reliability, relationship, 156–157tools, 162–164

Maintenance cost data collection, 85Maintenance cost estimation models,

83–85model I, 83model II, 84model III, 81model IV, 81–85

Maintenance costing, factors influencing, 78–79

Maintenance costs, factors influencing, 78

K10869_Book.indb 201 8/26/09 2:15:03 PM

202 Index

Maintenance labor cost estimation, 80–81corrective, 80–81preventive, 80–81

Maintenance manual, software life cycle cost element, 98

Maintenance material, 81–83Maintenance personnel skills,

experience, 78Manufacturing cost, categories, 72Manufacturing cost estimation models,

73–75model I, 73model II, 73model III, 74model IV, 74–75

Manufacturing costs, 72Manufacturing warranty costs, 86Mean time to failure, 147–149Mean time to repair, 3Microelectronics and Reliability, 3MIL-HDBK-276-1 (MC), “Life Cycle

Cost Model for Defense Material Systems Data Collection Workbook,” Department of Defense, 4

MIL-HDBK-259 (Navy) “Life Cycle Cost in Navy Acquisitions,” Department of Defense, 4

Mission time, 3Modifications, software life cycle cost

element, 98Motor life cycle cost estimation model,

124–125MTTR. See Mean time to repairMultiplicative models, software cost

estimation, 99–100

N

National Research Council, Transportation Research Board, 5

National Technical Information Center, 6

New aircraft system spares cost estimation model, 136–137

Nonrecurring cost, 2NTIS. See National Technical

Information Center

O

Offshore Technology Conference, proceedings, 4

Operating and support costas software life cycle cost, 97software life cycle cost element,

98Operational environment, maintenance

costing, 78Operator expertise, experience,

maintenance costing, 78Ownership cost, 2

P

Peterson, D.E., 5Petroleum and Chemical Industry

Conference, proceedings, 4Plant cost estimation model, 129–130Pressure vessel cost estimation model,

134–135Prevention, defined, 63–64Preventive maintenance labor cost

estimation, 80–81Proceedings of the Annual ISSAT

International Conference on Reliability and Quality in Design, proceedings, 4

Procurement, defined, 2Product usability, 87–88Product usability cost estimation,

87–88Program error cost estimation model,

132Program requirements, software life

cycle cost element, 98Program test, software life cycle cost

element, 98“Project-Oriented Life Cycle

Costing Workshop for Energy Conservation in Buildings,” 5

Proposals, incorporating into planning process for, 30–31

Pulp and Paper Industry Technical Conference, proceedings, 4

Pump life cycle cost estimation model, 125–126

K10869_Book.indb 202 8/26/09 2:15:04 PM

Index 203

Q

Quality cost classifications, distribution, industrial sector, 66–68

appraisal cost, 67external failure cost, 67–68internal failure cost, 67prevention cost, 67

Quality cost indexes, quality cost reduction, 68–69

Quality cost reduction, quality cost indexes, 68–69

Quality Engineering, 4

R

Rail International, 4Railway Gazette International, 4Rebitzer, G., 5Recurring cost, 2Redundancy, 3Regulatory controls, maintenance

costing, 78Reliability, 2

maintainability, relationship, 156–157

Reliability acquisition cost estimation model, 130

Reliability Analysis Center, 6Reliability and Maintainability

Symposium, proceedings, 4Reliability cost classification, 63–64Reliability Engineering and System

Safety, 4Reliability improvement warranty

costs, 86Reliability-related tasks, models,

estimation costs of, 64model I, 64–66model II, 65model III, 65model IV, 65model V, 66

Reliability Society, IEEE, 7Reliability tools, 162–164Repair cost, 2Repair manpower, 81–83Repair parts costs, 81–83

S

Safety cost estimation models, 70–72model I, 70model II, 70–71model III, 71model IV, 71–72

Satellite procurement cost estimation model, 137

Seldon, M.R., 5Ships, operating/support costs, 111–112

formula I, 111formula II, 111formula III, 111–112formula IV, 112

Simple interest, 11–12Single payment future worth

formula, 15Single payment present value formula,

15–16Single-satellite system launch cost

estimation model, 137Society of Manufacturing Engineers, 7Software cost estimation

analytic models, 99composite models, 98–99linear models, 99methods, 97–103models, 97–103multiplicative models, 99–100tabular models, 98

Software costing, 95–96Software life cycle cost

elements, 97influencing factors, 96–97model, 96–97

Software life cycle cost element, subelements, 98

SOLE-International Society of Logistics, 7

Spare parts costs, 81–83Specific life cycle cost models, 50–55

model I, 50–51model II, 51–52model III, 52–53model IV, 53–54model V, 55

Steel structure life cycle cost, 118–119Storage tank cost estimation model, 134

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204 Index

Straight-line method, 22Structural Engineering Institute of the

American Society of Civil Engineers, 5

Sum-of-years-digits method, 21SYD method. See Sum-of-years-digits

methodSystem availability, unavailability,

160–162System integration, software life cycle

cost element, 98System requirements, software life

cycle cost element, 98

T

Tabular models, software cost estimation, 98

Tank gun system life cycle cost estimation model, 138–139

Technical proposals, 30–31Technical reports, manuals, 4–5Technical Services Department

American Society for Quality, 6Test and integration cost


Test procedures, software life cycle cost element, 98

Test revisions, software life cycle cost element, 98

Transportation Research Board, National Research Council, 5

Transportation Research Record, 4Transportation system life cycle

costing, 105–116aircraft airframe maintenance cost

drivers, 109–110aircraft cost drivers, 108–110aircraft life cycle cost, 105–107aircraft turbine engine life cycle

cost, 108car life cycle cost, 113–114cargo ship life cycle cost, 110–111categories of typical aircraft

manufacturing cost drivers, 109

city bus life cycle cost estimation model, 114

combat aircraft hydraulic, fuel systems cost drivers, 110

helicopter maintenance cost drivers, 09

operating/support costs for ships, 111–112

formula I, 111formula II, 111formula III, 111–112formula IV, 112

urban rail life cycle cost, 112Type of service, maintenance costing, 78Types of life cycle cost models, inputs,

43–44

U

Uniform periodic payment future amount formula, 16–17

Uniform periodic payment present value formula, 18–19

Urban and Regional Information Systems Association, conference proceedings, 4

Urban rail life cycle cost, 112Usability costing, 87User manual, software life cycle cost

element, 98

V

Validation cost, software life cycle cost element, 98

Value of annuity payments, with annuity present, future values, 19–20

Verification cost, software life cycle cost element, 98

W

Waste treatment facilities life cycle costs, 119–120

Weather radar life cycle cost estimation model, 139–141

Y

Yanuck, R.R., 5

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life cycle costing for engineersviii contents 3.3 life cycle costing application areas 28 3.4 life...

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